$$\sum_{n_{1}+n_{2}+\cdots+n_{k}=n}\binom{n}{n_{1}, n_{2}, \dotsc,n_{k}} \cdot n_{1} n_{2} \cdots n_{k}=\frac{n !}{(n-k) !} \cdot k^{n-k}$$
I try this but I don't know if I am right:
\begin{align} &\sum_{n_{1}+n_{2}+\cdots+n_{k}=n}\binom{n}{n_{1}, n_{2}, \dotsc,n_{k}} \cdot n_{1} n_{2} \cdots n_{k}\\ &=\sum_{n_{1}+n_{2}+\cdots+n_{k}=n,}\binom{n}{n_{1}, n_{2}, \dotsc,n_{k}} \cdot n_{1} n_{2} \cdots n_{k} \\ &=\sum_{n_{1}+n_{2}+\cdots+n_{n}=n,} \frac{n !}{\left(n_{1}-1\right) !\left(n_{2}-1\right) ! \cdots\left(n_{k}-1\right) !} \\ &=\sum_{n_{1}+n_{2}+\cdots+n_{k}=n,\left(n_{1}, i=1,2, \dotsc, k\right.} \frac{n(n-1) \cdots(n-k+1)(n-k) !}{\left(n_{1}-1\right) !\left(n_{2}-1\right) ! \cdots\left(n_{k}-1\right) !} \\ &=n(n-1) \cdots(n-k+1) \sum_{n_{i}+n_{2}+\cdots+n_{k}=n, k} \frac{(n-k) !}{\left(n_{1}-1\right) !\left(n_{2}-1\right) ! \cdots\left(n_{k}-1\right) !} \\ &=\frac{n !}{(n-k) !} \sum_{m_{1}+m_{2}+\cdots+m_{k}=n-k} \frac{(n-k) !}{m_{1} ! m_{2} ! \cdots m_{k} !} \\ &=\frac{n !}{(n-k) !} \sum_{m_{1}+m_{2}+\cdots+m_{k}=n-k}\binom{n-k}{m_{1}, m_{2}, \dotsc, m_{k}}\\ &=\frac{n !}{(n-k) !} \cdot k^{n-k} \end{align}
Yes, it’s correct, apart from the notational problem that Fawkes4494d3 noted in the comments and fixed. You could also prove the result combinatorially. Suppose that you have players numbered from $1$ through $n$ and teams numbered from $1$ through $k$. The lefthand side is the number of ways to assign the $n$ players to the $k$ teams and then choose one player on each team to be captain. Thus, it is the number of ways to assign the $n$ players to $k$ numbered teams and assign each team a captain.
The factor of $\frac{n!}{(n-k)!}$ on the righthand side is the number of ways of choosing a captain for each of the $k$ teams, and $k^{n-k}$ is the number of ways to assign each of the remaining $n-k$ players to one of the $k$ teams, so the righthand side is also the number of ways to assign the $n$ players to $k$ numbered teams and assign each team a captain.