Let $n$ be an integer greater than three. Prove that $1 !+2 !+\cdots+n !$ cannot be a perfect power.
Proof:
For $n=4,$ we have $1 !+2 !+3 !+4 !=33,$ which is not a perfect power. For $k \geq 5, k ! \equiv 0(\bmod 10) .$
It follows that for $n \geq 5$ $ 1 !+2 !+3 !+4 !+\cdots+n ! \equiv 3 \quad(\bmod 10) $ so it cannot be a perfect square, or an even power, for this reason.
For odd powers, the following argument settles all cases:
one checks the claim for $n<9$ directly; for $k \geq 9, k !$ is a multiple of $27,$ while $1 !+2 !+$ $\cdots+8 !$ is a multiple of $9,$ but not $27 .$
Hence $1 !+2 !+\cdots+n !$ cannot be a cube or higher power.
But if $1 !+2 !+$ $\cdots+8 !$ is a multiple of $9,$ but not $27 .$ then WHY it cannot be cube or higher power ???
This is confusing me for an hour , i am missing something , can someone pls tell me what is it?
thankyou
If $n^k$ is a multiple of a prime $p$, then $n$ must be a multiple of $p$ and hence $n^k$ a multiple of $p^k$. In particular, a cube or higher power is either a multiple of $27$ or not a multiple of $3$.