= Solution
For $n\ge2$, use the <binomial coefficient> identity $(n/k)\binom{n-1}{k-1}=\binom nk$. It reduces the left side to $\binom{n-1}{k-1}(k/n)^{m-1}$. Put $t=(k-1)/(n-1)$. Then $k/n=((n-1)t+1)/n$, and the <binomial theorem> gives
$$
\left(\frac kn\right)^{m-1}=\sum_{\ell=0}^{m-1}\binom{m-1}{\ell}\frac{(n-1)^{m-1-\ell}}{n^{m-1}}t^{m-1-\ell}.
$$
Thus the required coefficients are
$$
\boxed{a_{n,m,\ell}=\binom{m-1}{\ell}\frac{(n-1)^{m-1-\ell}}{n^{m-1}},\qquad0\le\ell\le m-1.}
$$
For $m=1$ the sum has the single coefficient $a_{n,1,0}=1$. For $k=1$, the power $t^0$ denotes the constant <monomial> $1$.
The original PDF allows $n=1$, but its displayed ratio $(k-1)/(n-1)$ is then $0/0$. Literally that expression is undefined. The meaningful endpoint identity is obtained before normalization:
$$
\binom nk\left(\frac kn\right)^m=\binom{n-1}{k-1}\frac1{n^{m-1}}\sum_{\ell=0}^{m-1}\binom{m-1}{\ell}(k-1)^{m-1-\ell}.
$$
This <polynomial> identity is valid at $n=k=1$, giving $1$ on both sides. The normalized formula and the <Bernstein monomial recurrence> below are used only for $n\ge2$.
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