I've recently managed to prove the following result and was hoping to know if it already exists? $ \def\lf{\left\lfloor} \def\rf{\right\rfloor}$
$$ \ln(n!) = \sum_{k=1}^{p_k < n}\left( \sum_{r=1}^\infty r \lf \frac{n}{p_k^r} \rf\right)\ln p_k = \sum_{k=1}^{p_k < n} \lf \frac{n}{p_k} \rf \ln p_k + \ln{\lf \frac{n}{p_k} \rf}! $$
where $n$ is any integer and $p_k$ is the k'th prime.
Questions
Do these equations already exist? Can this be used to show any non-trivial result?
Your first equation is wrong. The correct exponent of prime $p$ in the prime factorization of $n!$ is given by Legendre's formula:
$$\nu_p(n!)=\sum_{r=1}^{\infty} \left\lfloor \frac{n}{p^r} \right\rfloor$$
Thus, taking the logarithm, and the sum on $p$ running on prime $p\leq n$,
$$\log(n!)=\sum_{p\leq n}\left(\sum_{r=1}^{\infty} \left\lfloor \frac{n}{p^r}\right\rfloor\right)\log(p)$$
Your second equation is also wrong. For instance,
$$\log(5!)=\log120 \neq \sum_{k=1}^{p_k < n} \lfloor \frac{n}{p_k} \rfloor \ln p_k + \ln{\lfloor \frac{n}{p_k} \rfloor}!=[2\log2+\log(2!)]+[\log3+\log(1!)]=\log24$$