Question: Let $R$ be a noetherian ring and $M$ be a finitely generated $R$ - Module.

Prove that there exists $n\in \mathbb N$ such that there is a finitely generated $R$ sub-module $N\subset R^n$ such that 

$$R^n / N \cong M$$

Any ideas ? Maybe use the isomorphism theorem
?

Answer: 
$$\frac{(2n)!-2(n!)^2}{(n!)^2 n}
=
\frac{1}{n} \left( \frac{(2n)!}{(n!)^2} - 2 \right)
=
\frac{1}{n} \left( \binom{2 n}{n} - 2 \right).$$

So one has to prove that if $n$ is prime, then 
$$
\binom{2 n}{n} \equiv 2 \pmod{n}.
$$
This is a very special case of [Lucas's Theorem](https://en.wikipedia.org/wiki/Lucas's_theorem). In this particular instance, just [compute](https://en.wikipedia.org/wiki/Frobenius_endomorphism) in $\mathbb{Z}/ n\mathbb{Z} [x]$
$$
\sum\_{i=0}^{2n} \binom{2n}{i} x^{i} = (1 + x)^{2n} = (1 + x^{n})^{2} = 1 + 2 x^{n} + x^{2 n}
$$
and compare the coefficients of $x^{n}$.
That is a consequence of [Wolstenholme's theorem](https://en.wikipedia.org/wiki/Wolstenholme's_theorem)
$$ \binom{2p-1}{p-1}\equiv 1\pmod{p^3},\qquad \binom{2p}{p}\equiv 2\pmod{p} $$
holding for any prime $p>3$.