Show that $5^n\mid2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$. Show that $5^n\mid 2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$.
I tried using Euler's Theorem: $2^{4\cdot5^{n-1}} \equiv 1 \mod{5^n}$ and $3^{4\cdot5^{n-1}}  \equiv 1 \mod{5^n}$ but I can't find a way to use it.
I also tried showing that the sum of the remainders of $2^{5^{n-1}}+3^{5^{n-1}}$ is a multiple of $5^n$ with induction but got stuck again.
How do I prove this kind of statement?

Simply induct on $n$. Base cases are clear. Suppose, for $k=n-1$, $5^k \mid 2^{5^{k-1}}+3^{5^{k-1}}$. The goal is to prove now the claim for $k=n$. Let $a=2^{5^{k-1}}$ and $b=3^{5^{k-1}}$. What we want to prove is, $5^n\mid a^5+b^5$, provided $5^{n-1}\mid a+b$. Now, using $a^5+b^5=(a+b)(a^4-a^3 b +a^2 b^2 -ab^3+b^4)$, and the fact that, $a\equiv -b\pmod{5}$, it follows that, $a^4-a^3 b +a^2 b^2 -ab^3+b^4\equiv 5b^4\equiv 0\pmod{5}$. Since $5^{n-1}\mid a+b$, their product is therefore divisible by $5^n$, as desired.
