TITLE: German tank problem, simple derivation
QUESTION [6 upvotes]: I was reading the recent question on the German tank problem, and had trouble with one of the derivations in this section.

$$\sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom N k} = \frac{k(N+1)}{k+1}$$

I understand where the left-hand side comes from, in the context of the problem. I'm just having trouble arriving at the right-hand side.
\begin{align*}
\sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom N k} &= \sum_{m=k}^N m \frac{(m-1)!}{(k-1)!(m-k)!} \frac{k! (N-k)!}{N!}\\
&= k \sum_{m=k}^N \frac{m!}{(m-k)!} \frac{(N-k)!}{N!}
\end{align*}
This looks promising, since I have found the $k$ that appears in the right-hand side, but I don't know what to do with the sum.

REPLY [2 votes]: $$\sum_{m=k}^N m\frac{{m-1 \choose k-1}}{{N \choose k}}=\frac{1}{{N\choose k}}\sum_{m=k}^N m{m-1 \choose k-1}=\frac{1}{{N\choose k}}\sum_{m=k}^{N}k{m \choose k}$$
The last step uses an identity.
Now this equals $$\frac{k}{{N\choose k}}\sum_{m=k}^{N} {m \choose k}=\frac{k}{{N\choose k}}\sum_{m=0}^{N} {m \choose k} =\frac{k}{{N\choose k}}{N+1\choose k+1}$$