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?
 A: 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.
A: Sketch:
Consider $(2+3)^{5^{n-1}}=5^{5^{n-1}}.$  It is straightforward to show that $5^n$ divides this expression (since $n\leq 5^{n-1}$).
On the other hand, 
$$
(2+3)^{5^{n-1}}=\sum_{i=0}^{5^{n-1}}\binom{5^{n-1}}{i}2^i3^{5^{n-1}-i}=\left(2^{5^{n-1}}+3^{5^{n-1}}\right)+\sum_{i=1}^{5^{n-1}-1}\binom{5^{n-1}}{i}2^i3^{5^{n-1}-i}.
$$
Now, all that must be shown is that the remaining sum is divisible by $5^n$.  
Now, count the number of factors of $5$ in $\binom{5^{n-1}}{i}$.  For $5^{n-1}!$, you have $\sum_{j=1}^{n-1}\frac{5^{n-1}}{5^j}$ factors of $5$ in the product.  This is a geometric sum, which you can compute explicitly.  Do something similar for the denominator.
A: Hint:
If $a+b=c5^r$ where $a,b,c$ are integers
$$(a+b)^5=a^5+b^5+\binom51ab(a^3+b^3)+\binom52a^2b^2(a+b)$$
$$\implies a^5+b^5=(a+b)^5-5ab(a^3+b^3)-10a^2b^2(a+b)$$ is clearly divisible by $5^{r+1}$
Set $a=2,b=5d-2,$ where integer $d$ is not divisible by $5$  then $r=1$
By induction, we can establish the proposition
