Let $N$ a positive integer. Denote $\mathcal{P}$ the set of prime numbers. I have to show that
\begin{align}
\log(N!) = \sum_{p^{\nu}\leq N \\ p\in \mathcal{P}} \left\lfloor\dfrac{N}{p^{\nu}}\right\rfloor \cdot \log(p)
\end{align}
To do it, I'd like to prove a more general statement, which says
\begin{align}
\sum_{n\leq x} \log(n) = \sum_{p^{\nu}\leq x \\ p\in \mathcal{P}} \left\lfloor\dfrac{x}{p^{\nu}}\right\rfloor \cdot \log(p) \qquad \forall\, x\geq1
\end{align}
$\textbf{My attempt}$:
I tried to using the Fundamental Theorem of Arithmetics to write every $n\in \mathbb{N}^{*}$ as product of primes, i.e.
\begin{align}
n= \prod_{p\in P \\
v\geq 0} p^{\nu} \
\end{align}
Then,
\begin{align}
\log(n) = \sum_{p\in \mathcal{P}\\ v\geq 0} \nu \cdot \log(p) \qquad (\ast)
\end{align}
Now I have some problem to rewrite $\sum_{n\leq x} \log(n)$ using $(\ast)$. I tried to find an equivalent condition to $n\leq x$ whithout success.
Any suggestions? Thanks in advance!
Best Answer
Hint: Instead of taking each integer of $N!=\prod_{n=1}^N n$ separately, take them all together and consider each prime on the entire factorial. Specifically, show that the largest $E$ such that $p^E\mid N!$ is $$ \sum_{\nu=1}^\infty\left\lfloor\frac N{p^\nu}\right\rfloor $$