Why does this expected value simplify as shown? I was reading about the german tank problem and they say that in a sample of size $k$, from a population of integers from $1,\ldots,N$ the probability that the sample maximum equals $m$ is:
$$\frac{\binom{m-1}{k-1}}{\binom{N}{k}}$$
This make sense. But then they take expected value of the sample maximum and claim:
$$\mu = \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom{N}{k}} = \frac{k(N+1)}{k+1}$$
And I don't quite see how to simplify that summation. I can pull out the denominator and a $(k-1)!$ term out and get:
$$\mu = \frac{(k-1)!}{\binom{N}{k}} \sum_{m=k}^N m(m-1) \ldots (m-k+1)$$
But I get stuck there...

Call $B_k^N=\sum\limits_{m=k}^N\binom{m-1}{k-1}$. 


Fact 1: $B_k^N=\binom{N}{k}$ (because the sum of the masses of a discrete probability measure is $1$ or by a direct computation).



Fact 2: For every $n\geqslant i\geqslant 1$, $n\binom{n-1}{i-1}=i\binom{n}{i}$.


Now to the proof.
Fact 2 for $(n,i)=(m,k)$ gives $\sum\limits_{m=k}^Nm\binom{m-1}{k-1}=\sum\limits_{m=k}^Nk\binom{m}{k}=k\sum\limits_{m=k+1}^{N+1}\binom{m-1}{(k+1)-1}=kB_{k+1}^{N+1}$. 
Fact 1 gives $B_{k+1}^{N+1}=\binom{N+1}{k+1}$. 
Fact 2 for $(n,i)=(N+1,k+1)$ (or a direct computation) gives $(k+1)B_{k+1}^{N+1}=(N+1)B_k^N$.
Finally, $\mu=\dfrac{kB_{k+1}^{N+1}}{B_k^N}=k\dfrac{N+1}{k+1}$.

Edit The same method yields, for every $i\geqslant0$,
$$
\mathrm E(X(X+1)\cdots(X+i))=\frac{k}{k+i+1}(N+1)(N+2)\cdots(N+i+1).
$$
