Alternative factorization of $\prod\limits^{n}_{k=1}k!^{k+1}$

Question

Question: How can I succinctly express (using the product and sum notations) the following expression? $$n^{(n+1)}(n-1)^{(n+1)+n}(n-2)^{(n+1)+n+(n-1)}\cdot\cdot\cdot 1^{(n+1)+n+(n-1)+\cdot\cdot\cdot+2}$$

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Fun fact: The following is in fact alternatively expressible as: $$\prod^{n}_{k=1}k!^{k+1}=\prod^{n}_{k=1}sf(k)H(k) \hspace{1mm}\rightarrow \hspace{3mm}sf(k)=\prod^{k}_{i=1}i!\hspace{1mm};H(k)=\prod^{k}_{i=1}i^i$$

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EDIT: I think I got it: The given expression equals:$$\prod^{n}_{k=1}k^{\sum^{(n+1)}_{i=1}-\sum^{k}_{i=1}}=\prod^{n}_{k=1}k^{\frac{1}{2}[(n+1)(n+2)-k(k+1)]}$$

Answer 1

Your representation is nice and correct. Here is a derivation

\begin{align*} \prod_{k=1}^n{k!}^{k+1}&=\prod_{k=1}^n\prod_{j=1}^kj^{k+1}\\ &=\prod_{j=1}^n\prod_{k=j}^{n}j^{k+1}\tag{1}\\ &=\prod_{j=1}^n{j}^{\sum_{k=j}^n(k+1)}\\ &=\prod_{j=1}^n{j}^{\left(\frac{n(n+1)}{2}-\frac{(j-1)j}{2}\right)+\left(n-(j-1)\right)}\\ &=\prod_{j=1}^nj^{\frac{1}{2}[(n+1)(n+2)-j(j+1)]} \end{align*}

Comment:

• In (1) we exchange the products. Note, the index range is $1\leq j\leq k\leq n$.