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 $x^{n}$.
Count the number of ways to arrange $n$ red and $n$ green balls into a sequence of $2n$ balls; there are $2n\choose n$ such arrangements. 
Say that two such arrangements are equivalent, if they can be transformed into another by a *cyclic* permutaion of the *first* $n$ balls.
If $n$ is prime, we have two cases:

* First all red, then all green balls or vice versa. These arrangements are only equivalent to themselves.
* At least one green and one red ball in the first half. Then all $n$ cyclic rearrangements are different (the first repetition must occur at a divisor of $n$)

Thus the $2n\choose n$ arrangements are partitioned into $2$ arrangements of the first kind and some integer $k$ number of equivalence classes containing (if $n$ is prime) $n$ elements each.
Hence 
$${2n\choose n}=2+kn, $$
or solved for $k$,
$$k=\frac{{2n\choose n}-2}n=\frac{\frac{(2n)!}{n!n!}-2}{n}=\frac{(2n)!-2n!^2}{n!^2n}=a\_n $$