Can the sum of $2025$ consecutive factorials be a perfect power?
My thoughts:
If the sum of $2025$ consecutive factorials is a perfect power, then the number of trailing zeroes in the sum must be a perfect (number of trailing zeroes)th power.
So this means that if the sum of $2025$ consecutive factorials is a perfect power, then the sum must be a perfect (number of trailing zeores)th residue $\pmod{p}$, where $p$ is a certain prime number.
For example, the sum of $10!+11!+12!+\cdots+2034!$ contains $2$ trailing zeroes as all factorials from $10!$ to $14!$ contains $2$ trailing zeroes, and since the sum contains $2$ trailing zeroes, then it must be a perfect square in order to be a perfect power.
By using Pari GP, I tried to check all factorials from $1!$ to $10^{6}!$, but I did not succeed.
Best Answer
Suppose $X=n!+(n+1)!+\cdots+(n+2024)!$ is a perfect power. Let $$Y=\frac X{n!}=1+(n+1)+(n+1)(n+2)+\cdots+(n+1)\cdots(n+2024).$$ Note that $$Y<\frac{(n+2025)!}{n!}<(n+2025)^{2025}.$$ For $X$ to be a perfect power, each prime which divides $n!$ exactly once must also divide $Y$. In particular, $$(n/2)^{\pi(n)-\pi(n/2)}\leq \prod_{n/2<p\leq n}p\leq Y\leq (n+2025)^{2025}.$$ This implies that $$\pi(n)-\pi(n/2)\leq 2025\log_{n/2}(n+2025);$$ as long as $n>200$, say, $n+2025<(n/2)^2$, and so we have \begin{equation}\tag{$\star$}\pi(n)-\pi(n/2)\leq 4050.\end{equation} On the other hand, we have explicit bounds on $\pi(n)$ from explicit versions of the prime number theorem: see for example this article of Dusart, as described on Wikipedia, which gives $$\frac{x}{\log x-1}\leq \pi(x)\leq \frac{x}{\log x-1.1}$$ for $x\geq 60184$. This implies, for $n\geq 130000$, \begin{align} \pi(n)-\pi(n/2) &\geq \frac n{\log n-1}-\frac{n/2}{\log(n/2)-1.1}\\ &>\frac n{\log n-1}\left(1-\frac{\log n-1}{2(\log(n/2)-2)}\right)\\ &=\frac n{\log n-1}\cdot\frac{\log n-3}{2\log n-4}. \end{align} As long as $\log n>10$, we have $$\pi(n)-\pi(n/2)\geq \frac n{\log n-1}\cdot\frac 7{16}.$$ Since $n/(\log n-1)$ is an increasing function of $n$, we have that, when $n\geq 130000$, $$\pi(n)-\pi(n/2)\geq \frac 7{16}\frac{130000}{\log(130000)-1}\geq \frac{7}{16}\cdot\frac{130000}{11}\geq 5000.$$ This contradicts ($\star$). We conclude that $X$ can only be a perfect power for $n<130000$. Since you have already checked these cases, we conclude that the sum in question is never a perfect power.