Is $1^{1!}+2^{2!}+3^{3!}+\dots n^{n!}$ a perfect square?
Obviously, $1$ is a perfect square. When $n=2$, the sum is $5$, which is not a perfect square. When $n=3,4$, the sum is $2\pmod{4}$. For $n=5$, the last 2 digits of the sum is $15$. When $n=6$, the sum is $7\pmod{8}$. When $n=7,8$, the number is either $2,3,7,8\pmod{10}$. But when $n=9$, the sum is $1\pmod{8}$, the sum of digits of divisible by $9$, and the last digit is $9$, but still not a perfect square.
Other than $1$, are there any perfect square in the sequence $1^{1!}+2^{2!}+3^{3!}+\dots n^{n!}$?
I think $1^{1!}+2^{2!}+3^{3!}+\dots n^{n!}$ will be a perfect square number if $n\equiv\pm1\pmod{10}$.
Best Answer
Hint: $n^{n!}$ is a perfect square (when $n\ge2$). What is the next perfect square after $n^{n!}$? How large is it compared to the quantity in the OP?