Let $X\sim HyperGeo(N,M,n)$ thus we get that: $$f_X(x)=\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}\;\ \text{with}\ x=max\{0,n-(N-M)\},...,min\{n,M\}$$
Then the expectation of $X$ is given by: $$\Bbb E(X)=\sum_{x=max\{0,n-(N-M)\}}^{min\{n,M\}}x\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}=\frac{1}{\binom{N}{n}}\sum_{x=max\{0,n-(N-M)\}}^{min\{n,M\}}x\binom{M}{x}\binom{N-M}{n-x}\\ =\frac{M}{\binom{N}{n}}\sum_{x=max\{0,n-(N-M)\}}^{min\{n,M\}}\binom{M-1}{x-1}\binom{N-M}{n-x}$$
So, since $x=max\{0,n-(N-M)\},...,min\{n,M\}$ we can make $x=1,...,(n-1)$ and clearly $f_X(x)$ has $0$ value for the $x$ values not met in its domain. So:
$$\Bbb E(X)=\frac{M}{\binom{N}{n}}\sum_{x=1}^{n-1}\binom{M-1}{x-1}\binom{N-M}{n-x}$$
And Vandermonde's identity says that: $$\binom{m+l}{r}=\sum_{k=0}^r\binom{m}{k}\binom{l}{r-k}$$
So let $r=n-1,\; k=x-1,\; m=M-1,\; l=N-M$, thus $r-k=n-x$. So, applying Vandermonde's identity we get that:
$$\Bbb E(X)=\frac{M}{\binom{N}{n}}\binom{N-1}{n-1}$$
and, since $\binom{N}{n}=\frac{N}{n}\binom{N-1}{n-1}$ we get that: $$\Bbb E(X)=\frac{nM}{N\binom{N-1}{n-1}}\binom{N-1}{n-1}=\frac{nM}{N}$$
Thus $$\Bbb E(X)=\frac{nM}{N}$$
