How do I make this formula for the primes more concise?

Question

The form I made for the $(n+1)^{th}$ prime $p_{n+1}$ is

$\displaystyle1+\sum_{j=1}^{2p_n-1}\lfloor\frac{p_n!^j}{j!}\rfloor-\lfloor\frac{p_n!^j-1}{j!}\rfloor=p_{n+1}.$

Problem is, just like any formula for $p_n,$ its very inefficient. To spite this, I like the form of it and I'm wondering if anyone can make it more concise?

Note: The sum is just summing $1's$ because the argument of the sum is just an indicator function that is $1$ if $j!|p_n!^j$ and $0$ otherwise. So it can be written like this,

$\displaystyle 1+\sum_{j=1}^{2p_n-1} 1_{j! | p_n!^j}=p_{n+1}$

So, manipulation of indicator functions of this form could help. Also, the sum only needs to be summed to $p_{n+1}-1$ to be correct, but the indicator function is $0$ for every value past that so I use Bertrand's postulate and sum to $2p_n-1$ to get rid of $p_{n+1}$ on the left side.

Answer 1

I'm using an answer to my own question to close this question because it doesn't seem that the formula can be made any simpler. If someone has an answer they are welcome to give it and I will probably accept it.