Factoring the factorials

Question

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).

Answer 1

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.

Answer 2

Perhaps consider this. Say you multiply $(n-1)!$ by $n$ to get $n!$. Then, if you continue multiplying by $n+1$ and $n+2$, etc, you won't multiply by another multiple of $n$ until you get to $2n$. $2n-n$ is n, so the $n^1$ factor will be show up $n$ times before being duplicated.