I've seen this stated in a few places.
If $$\vartheta(x) = \sum_{p\le{x}} \log (p) \qquad \psi(x) = \sum_{m=1}^{\infty}\vartheta\left(\sqrt[m]{x}\right)$$ Then $$\log(x!) = \sum_{m=1}^{\infty} \psi\left(\frac{x}{m}\right).$$
It is used by Ramanujan here. It is used by Jitsuro Nagura here.
Can anyone provide a proof for why it is true or provide a link to a proof?
Thanks very much.
We find
$$\sum_{m\ge1}\psi\left(\frac{x}{m}\right)=\sum_{m\ge1}\sum_{k\ge1}\vartheta\left(\sqrt[k]{\frac{x}{m}}\right)=\sum_{m\ge1}\sum_{k\ge1}\sum_{p\le \sqrt[k]{x/m}}\log p$$
$$=\sum_{m\ge1}\sum_{k\ge1}\sum_{mp^k\le x}\log p=\log \prod_{p\le x}p^{\#\{(m,k):mp^k\le x\}}=\log x!$$
since when counting $\#\{(m,k):mp^k\le x\}$, one sees for every $1\le n\le x$ there are $t=v_p(n)$ different tuples $(m,t),(mp,t-1),\cdots,(mp^{t-1},1)$ counted in the set (note $k\ge1$).