German tank problem, simple derivation

There is a binomial identity that $\displaystyle \sum_{m=k}^N \binom{m}{k} = \binom{N+1}{k+1}$,

You can see this by induction on $N$, if $N = k$, $\binom{k}{k} = \binom{k+1}{k+1}$, and if it is true that $$\displaystyle \sum_{m=k}^N \binom{m}{k} = \binom{N+1}{k+1}$$ then $$\displaystyle \sum_{m=k}^{N+1} \binom{m}{k} = \binom{N+1}{k+1} + \binom{N+1}{k} = \binom{N+2}{k+1}$$.

Now your sum was $$k \displaystyle \sum_{m=k}^N \frac{m!}{(m-k)!} \frac{(N-k)!}{N!} = \frac{k!k(N-k)!}{N!} \displaystyle \sum_{m=k}^N \binom{m}{k} = \frac{k(N-k)!k!}{N!} \binom{N+1}{k+1} = \frac{k(N+1)}{k+1}$$

$$\sum_{m=k}^N m\frac{{m-1 \choose k-1}}{{N \choose k}}=\frac{1}{{N\choose k}}\sum_{m=k}^N m{m-1 \choose k-1}=\frac{1}{{N\choose k}}\sum_{m=k}^{N}k{m \choose k}$$ The last step uses an identity.

Now this equals $$\frac{k}{{N\choose k}}\sum_{m=k}^{N} {m \choose k}=\frac{k}{{N\choose k}}\sum_{m=0}^{N} {m \choose k} =\frac{k}{{N\choose k}}{N+1\choose k+1}$$