This has an answer using Mellin transform methods. We use the material
from this MSE link
which is so very similar that we may use it as a template, making
adjustments as necessary.
Re-write your sum like this (here we have $a\gt 1$):
$$S_m(n) = \sum_{k=1}^n k^m a^k =
n^m \sum_{k=1}^n \left(\frac{k}{n}\right)^m a^k
\\ = n^m a^n + n^m \sum_{k=1}^{n-1} \left(\frac{k}{n}\right)^m a^k
= n^m a^n + n^m a^n \sum_{k=1}^{n-1}
\left(1 - \frac{k}{n}\right)^m a^{-k}.$$
Now consider the harmonic sum
$$S(x) = \sum_{k\ge 1} a^{-k} H_m(kx)$$
where $H_m(x)$ is the Heaviside step function defined by
$$H_0(x) = \begin{cases} & 1&\text{if}\quad x\in[0,1] \\
& 0 &\text{otherwise}\end{cases}
\quad\text{and}\quad H_m(x) = (1-x)^m H_0(x)
\quad\text{when}\quad m\in\mathbb{Z}^+.$$
We see that $$S(1/n) = \sum_{k=1}^n
a^{-k} \left(1 - \frac{k}{n}\right)^m,$$
i.e. the sum that remains to be computed.
But $S(x)$ is harmonic and may be evaluated by inverting its Mellin transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = a^{-k}, \quad \mu_k = k \quad \text{and} \quad
g(x) = H_m(x).$$
We need the Mellin transform $H_m^*(s)$ of $H_m(x)$ which is
$$\int_0^\infty H_m(x) x^{s-1} dx
= \frac{m!}{s(s+1)(s+2)\cdots(s+m)}$$
as is easily seen by induction using integration by parts.
It follows that the Mellin transform $Q(s)$ of $S(x)$ is given by
$$ Q(s) = \frac{m!}{s(s+1)(s+2)\cdots(s+m)} \mathrm{Li}_s(1/a)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \mathrm{Li}_s(1/a).$$
We thus obtain the Mellin inversion integral $$ S(x) = \frac{1}{2\pi
i} \int_{1/2-i\infty}^{1/2+i\infty} Q(s)/x^s ds$$ which we evaluate by
shifting it to the left for an expansion about zero. The fundamental
strip here is $\langle 0,+\infty \rangle$ which is determined by the
Heaviside step function term (the polylog converges everywhere).
For the pole at $s=0$ we get
$$\mathrm{Res}(Q(s)/x^s; s=0) =
\frac{m!}{m!} \mathrm{Li}_{0}(1/a) = \frac{1}{a-1}.$$
For the poles at $s=-1, -2, \ldots -m = -q$ we require
$$\mathrm{Li}_{-q}(z) =
\frac{1}{(1-z)^{q+1}} \sum_{p=0}^{q-1}
\left\langle {q\atop p} \right\rangle z^{q-p}.$$
Observe that all terms in the sum for $\mathrm{Li}_{-q}(1/a)$ are
positive (cite e.g. the combinatorial definition of the Eulerian
numbers) and hence none of the poles from the Heaviside term are
canceled. We may thus continue with
$$\mathrm{Res}(Q(s)/x^s; s=-q)
\\ = \left.\frac{m!}{s(s+1)\cdots(s+q-1)(s+q+1)\cdots(s+m)}
\frac{\mathrm{Li}_{s}(1/a)}{x^s}\right|_{s=-q}
\\= \frac{m!}{(-1)^q q!\times (m-q)!}
x^q \frac{1}{(1-1/a)^{q+1}} \sum_{p=0}^{q-1}
\left\langle {q\atop p} \right\rangle (1/a)^{q-p}
\\ = (-1)^{q} {m\choose q}
x^q \frac{1}{(a-1)^{q+1}} \sum_{p=0}^{q-1}
\left\langle {q\atop p} \right\rangle a^{p+1}.$$
Now returning to the sum and using $S(1/n)$ we finally have
$$S_m(n) = n^m a^n +
n^m a^n \left(\frac{1}{a-1}
+ \sum_{q=1}^m (-1)^{q} {m\choose q}
n^{-q} \frac{1}{(a-1)^{q+1}} \sum_{p=0}^{q-1}
\left\langle {q\atop p} \right\rangle a^{p+1}\right).$$
This is
$$n^m \frac{a^{n+1}}{a-1}
- \sum_{q=1}^m n^{m-q} {m\choose q}
\frac{1}{(1-a)^{q+1}} \sum_{p=0}^{q-1}
\left\langle {q\atop p} \right\rangle a^{n+p+1}.$$
We discover upon verification of this formula that it is missing a
constant term that does not depend on $n,$ which indicates that the
remainder integral does not vanish. To compute the constant we require
$$\sum_{q=1}^m {m\choose q}
\frac{1}{(1-a)^{q}} \sum_{p=0}^{q-1}
\left\langle {q\atop p} \right\rangle a^{p}
\\ = \frac{1}{(1-a)^m} \sum_{q=1}^m {m\choose q}
(1-a)^{m-q} \sum_{p=0}^{q}
\left\langle {q\atop p} \right\rangle a^{p}.$$
What we have here is a convolution of two EGFS that yield
$$m! (1-a)^{-m} [z^m]
\exp((1-a)z) \left(-1 + \frac{a-1}{a-\exp((a-1)z)}\right)
\\ = m! [z^m] \exp(z) \left(-1 + \frac{a-1}{a-\exp(-z)}\right)
= m! [z^m] \frac{1-\exp(z)}{a-\exp(-z)}.$$
Collecting everything we find
$$\frac{a^2}{1-a} + a +
\frac{a^2}{1-a} m! [z^m] \frac{1-\exp(z)}{a-\exp(-z)}
\\ = \frac{a^2}{1-a}
\left(1 + \frac{1-a}{a}
+ m! [z^m] \frac{1-\exp(z)}{a-\exp(-z)} \right)
\\ = \frac{a^2}{1-a}
m! [z^m] \left(\frac{1}{a} \exp(z)
+ \frac{1-\exp(z)}{a-\exp(-z)} \right)
\\ = \frac{a^2}{1-a}
m! [z^m] \frac{-1/a + 1}{a-\exp(-z)}
= \frac{a}{1-a}
m! [z^m] \frac{a-1}{a-\exp(-z)}
\\ = \frac{a}{(1-a)^{m+1}}
m! [z^m] \frac{a-1}{a-\exp((a-1)z)}
= \frac{a}{(1-a)^{m+1}}
\sum_{p=0}^{m-1} \left\langle {m\atop p} \right\rangle a^{p}.$$
We have established the following conjecture for $S_m(n):$
$$\bbox[5px,border:2px solid #00A000]{
\color{blue}{n}^m \frac{a^{n+1}}{a-1}
- \sum_{q=1}^m \color{blue}{n}^{m-q} {m\choose q}
\frac{1}{(1-a)^{q+1}} \sum_{p=0}^{q-1}
\left\langle {q\atop p} \right\rangle a^{n+p+1}
+ \frac{a}{(1-a)^{m+1}}
\sum_{p=0}^{m-1} \left\langle {m\atop p} \right\rangle a^{p}.}$$
This is a finite expansion in $n$ and $a$ with $m+2$ terms. E.g. we obtain
for $m=5$ and $a=11:$
$${\frac {11\,{n}^{5}{11}^{n}}{10}}-{\frac {11\,{n}^{4}{11}^{n}}{20}}
+{\frac {33\,{n}^{3}{11}^{n}}{25}}-{\frac {913\,{n}^{2}{11}^{n}}{500}}
+{\frac {957\,n{11}^{n}}{625}}-{\frac {7909\,{11}^{n}}{12500}}
+{\frac {7909}{12500}}.$$
Addendum. We also have from first principles that
$$S_m(n) = [z^n] \frac{1}{1-z} \mathrm{Li}_{-m}(az)
= [z^n] \frac{1}{1-z} \frac{1}{(1-az)^{m+1}} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle a^{m-p} z^{m-p}.$$
This is
$$(-1)^m \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(z-1/a)^{m+1}} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle \frac{1}{a^{p+1}} z^{m-p}.$$
Residues sum to zero and the residue at infinity is zero by
inspection. We get from the residue at $z=1$ (flip sign)
$$(-1)^{m+1} \frac{1}{(1-1/a)^{m+1}} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle \frac{1}{a^{p+1}}
= (-1)^{m+1} \frac{a}{(a-1)^{m+1}} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle a^{m-1-p}
\\ = \frac{a}{(1-a)^{m+1}} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle a^p.$$
For the residue at $z=1/a$ we write (flip sign)
$$(-1)^{m+1} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle \frac{1}{a^{p+1}}
\mathrm{Res}_{z=0}
\frac{1}{z-1} \frac{1}{(z-1/a)^{m+1}} z^{m-p-n-1}.$$
We require the derivative
$$\frac{1}{m!} \left(\frac{1}{z-1} z^{m-p-n-1}\right)^{(m)}
\\ = \frac{1}{m!} \sum_{q=0}^m {m\choose q} \frac{(-1)^q q!}{(z-1)^{q+1}}
(m-p-n-1)^{\underline{m-q}} z^{m-p-n-1-(m-q)}
\\ = \sum_{q=0}^m \frac{(-1)^q}{(z-1)^{q+1}}
{m-p-n-1\choose m-q} z^{q-p-n-1}.$$
Evaluate at $z=1/a$ to get
$$(-1)^{m+1} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle \frac{1}{a^{p+1}}
\sum_{q=0}^m \frac{(-1)^q}{(1/a-1)^{q+1}}
{m-p-n-1\choose m-q} \frac{1}{a^{q-p-n-1}}
\\ = (-1)^{m+1} a^{n+1} \sum_{p=0}^{m-1}
\left\langle {m\atop p} \right\rangle
\sum_{q=0}^m \frac{(-1)^q}{(1-a)^{q+1}}
{m-p-n-1\choose m-q}.$$
With some algebra we have obtained the alternate closed form
$$\bbox[5px,border:2px solid #00A000]{
(-1)^{m+1} a^{n+1}
\sum_{q=0}^m \frac{(-1)^q}{(1-a)^{q+1}}
\sum_{p=0}^{m-1} \left\langle {m\atop p} \right\rangle
{p-n\choose m-q}
+ \frac{a}{(1-a)^{m+1}}
\sum_{p=0}^{m-1} \left\langle {m\atop p} \right\rangle a^{p}.}$$
Note that the binomial coefficient will produce polynomials in $n$ of
degree $m-q$ when we fix $m.$
