real analysis - A nice Combinatorial Identity

I am trying to show that $\forall N\in\mathbb{N}$,

$$\sum\limits_{n=0}^{N}\sum\limits_{k=0}^{N}\frac{\left(-1\right)^{n+k}}{n+k+1}{N\choose n}{N\choose k}{N+n\choose n}{N+k\choose k}=\frac{1}{2N+1}$$

It's backed by numerical verifications, but I can't come up with a proof.

So far, I tried using the generating function of $\left(\frac{1}{2N+1}\right)_{N\in\mathbb{N}}$, which is $\frac{\arctan\left(\sqrt{x}\right)}{\sqrt{x}}$, by showing that the LHS has the same generating function, but this calculation doesn't seem to lead me anywhere...

Any suggestion ?

Edit: the comment of bof (below this question) actually leads to a very simple proof.

Indeed, from bof's comment we have that the LHS is equal to $$\int_{0}^{1}\left(\sum\limits_{k=0}^{N}(-1)^k{N\choose k}{N+k\choose k}x^k\right)^2dx$$

And we recognize here the shifted Legendre Polynomials $\widetilde{P_N}(x)=\displaystyle\sum\limits_{k=0}^{N}(-1)^k{N\choose k}{N+k\choose k}x^k$.

And we know that the shifted Legendre Polynomials form a family of orthogonal polynomials with respect to the inner product $\langle f|g\rangle=\displaystyle\int_{0}^{1}f(x)g(x)dx$, and that their squared norm with respect to this product is $\langle\widetilde{P_n}|\widetilde{P_n}\rangle=\frac{1}{2n+1}$;

so this basically provides the desired result immediately.