That is an amazing identity. Let’s rewrite it as $$ \prod_{j=0}^N \binom{N}{j} = \frac{(\prod_{k=1}^N k^k)^2}{(N!)^{N+1}}. $$
The hard part is finding this identity. Once found, it can be proved by induction on $N$. The key point is to realize binomial coefficients have the identity $$ \binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1} $$ when $a\geq b \geq 1$.
You can check the identity holds at $N=1$ (I checked it up to $N = 4$ to make sure it wasn’t mistyped). Assuming it holds at $N$, \begin{align*} \prod_{j=0}^{N+1} \binom{N+1}{j} & = \prod_{j=1}^{N+1} \binom{N+1}{j} \\ & = \prod_{j=1}^{N+1} \frac{N+1}{j}\binom{N}{j-1}\\ & = \frac{(N+1)^{N+1}}{(N+1)!}\prod_{j=0}^N\binom{N}{j}. \end{align*} By induction, that last product can be rewritten and we get \begin{align*} \prod_{j=0}^{N+1}\binom{N+1}{j} & = \frac{(N+1)^{N+1}}{(N+1)!} \frac{(\prod_{k=1}^N k^k)^2}{(N!)^{N+1}} \\ & = \frac{(N+1)^{2(N+1)}(\prod_{k=1}^N k^k)^2}{(N+1)^{N+1}(N+1)!(N!)^{N+1}}. \end{align*} The numerator is what we want and the denominator is $(N+1)!((N+1)!)^{N+1}$, which is $(N+1)!^{N+2}$ and that is also what we want.