probability - Expected value of MLE of SRSWOR($n$)

Suppose an SRSWOR of size $n$ has been drawn from a population labelled $1,2,..,N$, where the population size $N$ is unknown.
(a)Find the maximum likelihood estimator $N'$ of $N$.
(b)Find the pmf of $N'$
(c)Show that $E(\frac{n+1}{n}N'-1)=N$

We know that the selection probability is $\frac{1}{{N \choose n}}$.So, the MLE is the $n$th order statistic or the maximum $X_{(n)}$.It's pmf is also, $P[X_{(n)}=m]=\frac{{m \choose n}-{m-1 \choose n}}{{N \choose n}}$
But I cannot show the last part.Please help

Answer

Let's rewrite the distribution of $N'=X_{(n)}$ a bit more precisely:

$$P(N'=j)=\begin{cases}\dfrac{\binom{j-1}{n-1}}{\binom{N}{n}}&,\text{ if }j=n,n+1,\ldots,N \\ \\\quad 0&,\text{ otherwise }\end{cases}$$

Then we have

$$\begin{align} E(N')&=\sum_{j=n}^N jP(N'=j) \\&=\frac{n}{\binom{N}{n}}\sum_{j=n}^N \frac{j}{n}\binom{j-1}{n-1} \\&=\frac{n}{\binom{N}{n}}\sum_{j=n}^N \binom{j}{n} \end{align}$$

If we can use the fact that $$\sum_{j=n}^N P(N'=j)=1\implies \sum_{j=n}^N \binom{j-1}{n-1}=\binom{N}{n}$$

, it follows that $$E(N')=\frac{n}{\binom{N}{n}}\binom{N+1}{n+1}=\frac{n(N+1)}{n+1}$$

Or you can prove this identity directly. See this, this and this.