Has $$\sum_{j=1}^p j!=q^r$$ , where q,p,r are positive integers, and r > 1 , a solution ?
I solved partially, if r is even, then RHS is a perfect square, and there is no doubt in that. Therefore, the left side must be a perfect square as well. But for p>3, the last digit of LHS is 3 which is not the last digit of any square. Hence, p<4, and hence manually checking, only solutions are p =1,3. But i cannot generalize when r is odd. Any solution to the next part would be helpful!
Let $S(p)=\sum_{j=1}^pj!$ denote your sum.
Claim: For $p≥8$ $v_3(S(p))=2$
(Here, as usual, for a prime $q$ and natural number $n$, $v_q(n)$ denotes the order to which $q$ divides $n$. Thus, $v_3(18)=2$ for example.).
Pf: One computes that $S(8)=3^2\times 11\times 467$ and from there after you have at least three factors of $3$ in each summand.
It follows immediately that, at least for $p≥8$ your sum could not be any power of degree greater than $2$. As you have already addressed the case of squares, we are done (after a simple search for small $p$).