Just for the fun of it, I've started factoring $n!$ into its prime divisors, and this is what I got for $2\leq n\leq20$:
$$\begin{align}
2! &= 2^\color{red}{1} &S_e=1\\
3! &= 2^\color{red}{1} \cdot 3^\color{red}{1} &S_e=2\\
4! &= 2^3 \cdot 3^\color{red}{1} &S_e=4\\
5! &= 2^3 \cdot 3^\color{red}{1} \cdot 5^\color{red}{1} &S_e=5\\
6! &= 2^4 \cdot 3^2 \cdot 5^\color{red}{1} &S_e=7\\
7! &= 2^4 \cdot 3^5 \cdot 5^\color{red}{1} \cdot 7^\color{red}{1} &S_e=8\\
8! &= 2^7 \cdot 3^2 \cdot 5^\color{red}{1} \cdot 7^\color{red}{1} &S_e=11\\
9! &= 2^7 \cdot 3^4 \cdot 5^\color{red}{1} \cdot 7^\color{red}{1} &S_e=13\\
10! &= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7^\color{red}{1} &S_e=15\\
11! &= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7^\color{red}{1} \cdot 11^\color{red}{1} &S_e=16\\
12! &= 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7^\color{red}{1} \cdot 11^\color{red}{1} &S_e=19\\
13! &= 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7^\color{red}{1} \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} &S_e=20\\
14! &= 2^{11} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} &S_e=22\\
15! &= 2^{11} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} &S_e=24\\
16! &= 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} &S_e=28\\
17! &= 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} \cdot 17^\color{red}{1} &S_e=29\\
18! &= 2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} \cdot 17^\color{red}{1} &S_e=32\\
19! &= 2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} \cdot 17^\color{red}{1} \cdot 19^\color{red}{1} &S_e=33\\
20! &= 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11^\color{red}{1} \cdot 13^\color{red}{1} \cdot 17^\color{red}{1} \cdot 19^\color{red}{1} &S_e=36\\
\end{align}$$
where $S_e$ is the sum of the exponents. There are lots of interesting patterns to chase here, but I've focused on the prime factors with exponent $1$ (highlighted). My conjecture is,
In the sequence above, prime factor $p^1$ will appear $p$ times starting from $p!$.
Or, otherwise stated,
$\forall k = 2p-1$, $k!$ is the last term of the sequence $\{a_n\}=n!$ to have $p^1$ as one of its prime factors.
[I could not find any reasonable pattern concerning $p^2$.]
This might be some well-known fact, but as someone who hasn't seen prime factorization since elementary school I was wondering how this could be proven (or whether it is even right, although it looks like it is).
Best Answer
You're just summing up https://oeis.org/A001222. It might be easier to work with this sequence without taking a cumulative sum. Apparently A001222 is called $\Omega(n)$. You can research from there.