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That is an amazing identity. The hardest part is finding it. Once found, it can be proved by induction on $N$.

I suggest you rewrite the identity as $$ \prod_{j=0}^N \binom{N}{j} = \frac{(\prod_{k=1}^N k^k)^2}{(N!)^{N+1}}, $$ and this is the version I will use below.

Before starting a proof, the key point to be aware of is that binomial coefficients admit 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 a positive integer $N$, we have \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. QED

As long as you keep in mind that $\binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}$, I would consider the argument above to be a straightforward induction: no special tricks are needed. It is the kind of thing a reader can be expected to derive on their own when the journal is tight on space. So if nobody can point to a published proof and the journal does not let you cite this MO page, just say the identity can be proved by induction on $N$, keeping in mind that $\binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}$ if $a\geq b \geq 1$.

That is an amazing identity. The hardest part is finding it. Once found, it can be proved by induction on $N$.

I suggest you rewrite the identity as $$ \prod_{j=0}^N \binom{N}{j} = \frac{(\prod_{k=1}^N k^k)^2}{(N!)^{N+1}}, $$ and this is the version I will use below.

Before starting a proof, the key point to be aware of is that binomial coefficients admit 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 a positive integer $N$, we have \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. QED

As long as you keep in mind that $\binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}$, I would consider the argument above to be a straightforward induction: no special tricks are needed. It is the kind of thing a reader can be expected to derive on their own when the journal is tight on space. So if nobody can point to a published proof and the journal does not let you cite this MO page, just say the identity can be proved by induction on $N$ while keeping in mind that $\binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}$ if $a\geq b \geq 1$.

Remark. There are few places in math where I have seen $k^k$ naturally show up (forget tetration, please). Besides the identity above, I can think of its role in Stirling’s estimate for $k!$ and in the Gauss—Legendre multiplication formula (distribution relation) $$ G(z)G(z+1/k)\cdots G(z+(k-1)/k) = \sqrt{k}\frac{G(kz)}{k^{kz}}, $$ where $G(z) = \Gamma(z)/\sqrt{2\pi}$.

That is an amazing identity. The hardest part is finding it. Once found, it can be proved by induction on $N$.

Let’s rewrite it as $$ \prod_{j=0}^N \binom{N}{j} = \frac{(\prod_{k=1}^N k^k)^2}{(N!)^{N+1}}. $$

Before starting a proof, the key point to be aware of is that binomial coefficients admit 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 a positive integer $N$, we have \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.

That is an amazing identity. The hardest part is finding it. Once found, it can be proved by induction on $N$.

I suggest you rewrite the identity as $$ \prod_{j=0}^N \binom{N}{j} = \frac{(\prod_{k=1}^N k^k)^2}{(N!)^{N+1}}, $$ and this is the version I will use below.

Before starting a proof, the key point to be aware of is that binomial coefficients admit 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 a positive integer $N$, we have \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. QED

As long as you keep in mind that $\binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}$, I would consider the argument above to be a straightforward induction: no special tricks are needed. It is the kind of thing a reader can be expected to derive on their own when the journal is tight on space. So if nobody can point to a published proof and the journal does not let you cite this MO page, just say the identity can be proved by induction on $N$, keeping in mind that $\binom{a}{b} = \frac{a}{b}\binom{a-1}{b-1}$ if $a\geq b \geq 1$.

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.

That is an amazing identity. The hardest part is finding it. Once found, it can be proved by induction on $N$.

Let’s rewrite it as $$ \prod_{j=0}^N \binom{N}{j} = \frac{(\prod_{k=1}^N k^k)^2}{(N!)^{N+1}}. $$

Before starting a proof, the key point to be aware of is that binomial coefficients admit 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 a positive integer $N$, we have \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.