Understanding a very elementary property of factorials
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 $k\ge1$).
A slightly different variation based upon the prime factorisation of $x!$.
We obtain for integers $x>0$ \begin{align*} \color{blue}{\log x!}&=\log\prod_{p\leq x}p^{\left\lfloor\frac{x}{p}\right\rfloor+\left\lfloor\frac{x}{p^2}\right\rfloor+\cdots}\tag{1}\\ &=\log\prod_{p\leq x}p^{\sum_{m=1}^\infty\left\lfloor\frac{x}{p^m}\right\rfloor}\tag{2}\\ &=\sum_{p\leq x}\sum_{m=1}^\infty\log p\left\lfloor\frac{x}{p^m}\right\rfloor\tag{3}\\ &=\sum_{m=1}^\infty\sum_{p\leq \sqrt[m]{x}}\log p\left\lfloor\frac{x}{p^m}\right\rfloor\tag{4}\\ &=\sum_{m=1}^\infty\sum_{p\leq \sqrt[m]{x}}\log p\sum_{j=1}^{x/p^m}1\tag{5}\\ &=\sum_{m=1}^\infty\sum_{j=1}^x\sum_{p\leq \sqrt[m]{x}}\log p\tag{6}\\ &\,\,\color{blue}{=\sum_{j=1}^x\psi\left(\frac{x}{j}\right)} \end{align*} and the claim follows.
Comment:
In (1) we do the prime factorisation of $x!$ and observe that $\left\lfloor\frac{x}{p}\right\rfloor$ counts the numbers $\leq x$ which are a multiple of $p$. Since we also have to add $1$ for each number which is a multiple of $p^2$ we add $\left\lfloor\frac{x}{p^2}\right\rfloor$, etc.
In (2) we use the series notation. Note the series is finite.
In (3) we use properties of the logarithm.
In (4) we exchange the order of the series.
In (5) we write the factor $\left\lfloor\frac{x}{p^m}\right\rfloor$ as sum.
In (6) we exchange the order of the series again.