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.