Let, $$\Theta(n)=\sum_{p^{\alpha}\leq n} \ln p $$ be the second chebyshev theta function. Then is it true that, $$\ln x!=\sum_{k\geq 1}\Theta\left(\frac{x}{k}\right)$$
If yes how can I prove that?
MY ATTEMPT:-
$$\ln(x!)=\sum_{p\leq x}\alpha(p)\ln p$$ Where $\alpha(p)$ is the highest power of $p$ in the prime factorisation of $x!$
It's the possible expression I could think of for $ln(x!)$. I would appreciate any hint.