I know that $S_n:=1!^2+2!^2+3!^2+\dots+n!^2$ cannot be a perfect square because it is equal to $2\pmod{3}$ and it is never a perfect cube because it is equal to $5\pmod{9}$, but can $S_n$ be a higher odd perfect power?
Edit:
-
$S_n$ cannot be also a perfect 5th power when $n\geq10$, since $17$ is not a 5th power residue $\pmod{100}$.
-
$S_n$ is also never a perfect 11th power if $n\geq23$, since $8$ is not a 11th power residue $\pmod{23}$.
-
$S_n$ is not a perfect 7th power for all $n\geq29$ because $7$ is not a 7th power residue $\pmod{29}$.
-
$S_n$ is not a perfect 13th power if $n\geq52$, since $7$ is not 13th power residue $\pmod{53}$.
-
$S_n$ isn't a perfect 17th power when $n\geq137$, since $70$ is not a 17th power residue $\pmod{137}$.
-
$S_n$ isn't a perfect 19th power when $n\geq191$, since $44$ is not 19th power residue $\pmod{191}$.
-
$S_n$ is never a perfect 23rd power if $n\geq47$, since $30$ is not a 23rd power residue $\pmod{47}$.
-
$S_n$ is not a perfect 29th power for all $n\geq59$, since $5$ is not 29th power residue $\pmod{59}$.
-
If $n\geq310$, then $S_n$ is never a perfect 31st power, since $62$ is not a 31st power residue $\pmod{311}$.
-
$S_n$ is not a perfect 37th power for all $n\geq148$, since $88$ isn't a 37th power residue $\pmod{149}$.
-
$S_n$ cannot be a perfect 41st power for all $n\geq82$, because $S_{82}\equiv45\pmod{83}$, and since $x^{41}\equiv45\pmod{83}$ is not solvable, $45$ is not a 41st power residue $\pmod{83}$.
Best Answer
Partial answer
Note that, for $n\geq9$, it is easy to show that $S(n)$ always has $1817$ as last digits. This fact eliminates the posibility of $S(n)$ being some even perfect power. For odd integers, note that:
perfect powers of integers having $1$ as last digit also have $1$ as last digit, so $S(n)$ can not be a perfect power of some odd integer having $1$ as last digit.
perfect powers of integers having $3$ as last digit have $3,9,7,1$ as last digits, so $S(n)$ could be a perfect power of some odd integer having $3$ as last digit only for powers $P$ of the form $P^{4k+3}$. It would remain to show that there is not any power of some odd integer having $3$ as last digit of the form $P^{4k+3}$ that could end in $1817$.
perfect powers of integers having $5$ as last digit also have $5$ as last digit, so $S(n)$ can not be a perfect power of some odd integer having $5$ as last digit.
perfect powers of integers having $7$ as last digit have $7,9,3,1$ as last digits, so $S(n)$ could be a perfect power of some odd integer having $7$ as last digit only for powers $P$ of the form $P^{4k+1}$. It would remain to show that there is not any power of some odd integer having $7$ as last digit of the form $P^{4k+1}$ that could end in $1817$.
perfect powers of integers having $9$ as last digit have $9,1$ as last digits, so $S(n)$ can not be a perfect power of some odd integer having $9$ as last digit.
Indeed, the "fixed" last digits of $S(n)$ grow as $n$ grows, so if there exists a counterexample for the pending cases, maybe other ones with more fixed ending digits could work.
EDIT
Running a Python program, I have been checking all odd integers less than $10001$ having $3$ or $7$ as last digit, and powers of the form $4k+3$ or $4k+1$ respectively less than $10001$.
I have been able to check that there are both powers of odd integers having $3$ as last digit of the form $P^{4k+3}$, and powers of odd integers having $7$ as last digit of the form $P^{4k+3}$, that end in $1817$. For instance, $73^{443}$ or $137^{57}$, although the program showed at least fourteen counterexamples before I stopped it.
I have found the powers $137^{2057}$, $153^{1399}$, $297^{929}$, $313^{1587}$, and $473^{2403}$ before I stopped the program, all of them ending in $51817$, the last five digits of $S(n)$ for $n\geq 14$
I have found the power $777^{653}$ ending in $851817$, the last six digits of $S(n)$ for $n\geq 14$, before I stopped the program.
Therefore, although the suitable powers get scarcer, the "fixed" last digits approach does not seem so promising as it could, or at least some additional insight would be needed. It seems that the smallest powers ending in some given fixed digits of $S(n)$ are quite bigger than $S(n)$, so proving that this holds would be a possible line of attack.
EDIT 2
Noting that $\lim_{n\to\infty} \frac{n!^2}{S(n)}=1$, we have that $S(n)<n^{2n}$; hence, there is an important restriction on the size of the possible perfect powers. Proving a minimum size of the powers generating the "fixed" digits greater than $n^{2n}$ could be a way to solve the problem.
Note also that the number of "fixed" digits of $S(n)$ seems to be always near to $\frac{n}{2}$.