It appears we can give another derivation of the closed form by @vadim123 for the
sum $$q_n = \sum_{j=k}^n m^j {n\choose j} {j \brace k}$$
using the bivariate generating function of the Stirling numbers of the
second kind. This computation illustrates generating function techniques as presented in Wilf's generatingfunctionology as well as the technique of annihilating coefficient extractors.
Recall the species for set partitions which is
$$\mathfrak{P}(\mathcal{U} \mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1)).$$
Introduce the generating function
$$Q(z) = \sum_{n\ge k} q_n \frac{z^n}{n!}.$$
We thus have
$$Q(z) = \sum_{n\ge k} \frac{z^n}{n!}
\sum_{j=k}^n m^j {n\choose j} {j \brace k}.$$
Substitute $G(z, u)$ into the sum to get
$$Q(z) = \sum_{n\ge k} \frac{z^n}{n!}
\sum_{j=k}^n m^j {n\choose j}
j! [z^j] \frac{(\exp(z)-1)^k}{k!}
\\ = \sum_{j\ge k} m^j \left([z^j] \frac{(\exp(z)-1)^k}{k!}\right)
\sum_{n\ge j} j! \frac{z^n}{n!} {n\choose j}
\\ = \sum_{j\ge k} m^j \left([z^j] \frac{(\exp(z)-1)^k}{k!}\right)
\sum_{n\ge j} \frac{z^n}{(n-j)!}
\\= \sum_{j\ge k} m^j \left([z^j] \frac{(\exp(z)-1)^k}{k!}\right)
z^j \sum_{n\ge j} \frac{z^{n-j}}{(n-j)!}
\\ = \exp(z)
\sum_{j\ge k} m^j z^j \left([z^j] \frac{(\exp(z)-1)^k}{k!}\right).$$
Observe that the sum annihilates the coefficient extractor, producing
$$Q(z) = \exp(z)\frac{(\exp(mz)-1)^k}{k!}.$$
Extracting coefficients from $Q(z)$ we get
$$q_n = \frac{n!}{k!}
[z^n] \exp(z)
\sum_{q=0}^k {k\choose q} (-1)^{k-q} \exp(mqz)
\\ = \frac{n!}{k!}
[z^n] \sum_{q=0}^k {k\choose q} (-1)^{k-q} \exp((mq+1)z)
= \frac{n!}{k!}
\sum_{q=0}^k {k\choose q} (-1)^{k-q} \frac{(mq+1)^n}{n!}
\\ = \frac{1}{k!}
\sum_{q=0}^k {k\choose q} (-1)^{k-q} (mq+1)^n.$$
Note that when $m=1$ $Q(z)$ becomes
$$\exp(z)\frac{(\exp(z)-1)^k}{k!}
= \frac{(\exp(z)-1)^{k+1}}{k!}
+ \frac{(\exp(z)-1)^k}{k!}$$
so that
$$[z^n] Q(z) = (k+1){n\brace k+1} + {n\brace k}
= {n+1\brace k+1},$$
which can also be derived using a very simple combinatorial argument.
Addendum.
Here is another derivation of the formula for $Q(z).$
Observe that when we multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
(I have included this derivation in several of my posts.)
Now in the present case we have
$$A(z) = \sum_{j\ge k} {j\brace k} m^j \frac{z^j}{j!}
\quad\text{and}\quad
B(z) = \sum_{j\ge 0} \frac{z^j}{j!} = \exp(z).$$
Evidently $A(z)$ is just the exponential generating function for set
partitions into $k$ sets evaluated at $mz,$ so we get
$$A(z) = \frac{(\exp(mz)-1)^k}{k!}$$
and with $Q(z) = A(z) B(z)$ the formula for $Q(z)$ follows.
