TITLE: Combinatorial identity $\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}$
QUESTION [7 upvotes]: How to prove the combinatorial identity?
$$\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}$$
I took $\ln$ of the left hand side
$$\sum_{j=0}^n \ln(j^{n+1}) - \ln((j!)^2)$$
but not going anywhere from here. any help is welcome

REPLY [2 votes]: Here is another algebraic approach.

We obtain
\begin{align*}
\prod_{k=1}^n\binom{n}{k}&=\prod_{k=1}^n\frac{n(n-1)\cdots(n-k+1)}{k!}\\
&=\left(\prod_{k=1}^n\frac{1}{k!}\right)\left(\prod_{k=1}^n\prod_{j=n-k+1}^nj\right)\tag{1}\\
&=\left(\prod_{k=1}^n\frac{1}{k!}\right)\left(\prod_{k=1}^n\prod_{j=k}^nj\right)\tag{2} \\
&=\left(\prod_{k=1}^n\frac{1}{k!}\right)\left(\prod_{j=1}^n\prod_{k=1}^jj\right)\tag{3} \\
&=\left(\prod_{k=1}^n\frac{1}{k!}\right)\left(\prod_{j=1}^nj^j\right)\tag{4} \\
&=\prod_{k=1}^n\frac{k^k}{k!}
\end{align*}

Comment:

In (1) we separate products of numerator and denominator
In (2) we exchange the order of multiplication in the second double product by letting $k \longrightarrow n-k$
In (3) we exchange the products of the second double product by observing the index range is $1\leq k\leq j\leq n$
In (4) we use $\prod_{k=1}^jj=j^j$