Note that $\binom ni$ is a polynomial in $n$ of degree $i$. Thus, for any given polynomial $a_0+\dotsb+a_dn^d$, there is exactly one way to write it as a linear combination of binomial coefficients:
$$ \sum_{i=0}^d \binom ni b_i = a_0 + a_1 n + a_2 n^2 + \dotsb + a_d n^d $$
(Consider this as a linear system in the variables $b_i$; since each $\binom ni$ has degree $i$, the system's coefficient matrix is triangular and has nonzero diagonal entries; so the system has unique solutions.)
Note that on the LHS there, we wrote $\sum_{i=0}^d$, and worked in the vector space of polynomials of degree at most $d$. But we could also write $\sum_{i=0}^{d+1}$, set $a_{d+1}=0$, and do the same in the vector space of polynomials of degree at most $d+1$. We find that there is a unique way to write blah blah blah. But the solution in the at-most-degree-$d$ world extends to a solution in the at-most-degree-$(d+1)$ world, just by taking $b_{d+1}=0$; so this must be the unique solution in the at-most-degree-$(d+1)$ world.
So: if you write a polynomial of degree $d$ as a linear combination of binomial coefficients, the coefficients (in the linear combination) of the binomial coefficients $\binom{n}{d+1},\binom{n}{d+2},\binom{n}{d+3},\dotsc$ are all zero.
Now to your problem. Fix $k\ge 2$. Let $c_i$ be the unique coefficients such that
$$ \sum_{i=0}^n \binom ni c_i = n^{k-1} \qquad\text{for all $n\ge 0$.} $$
As discussed above, since the RHS has degree $k-1$, we have $c_i=0$ for all $i\ge k$. Applying binomial inversion, we get
$$ c_n = \sum_{i=0}^n \binom ni (-1)^{n+i} i^{k-1} \qquad\text{for all $n\ge 0$.} $$
Since $k\ge 2$, we have $0^{k-1}=0$, so the $i=0$ term of the RHS is zero; thus the RHS is $(-1)^n$ times your sum. Since $c_n=0$ for all $n\ge k$, the desired result is proved.
