Here is a result of mine that I really wish to be strictly proven (I need it as an intermediate lemma for a theorem).
Let $v_5(a)$ indicate the $5$-adic valuation of $a \in \mathbb{Z}^+$.
How can we prove that $v_5((10^{k+t}+10^t+1)^c-1)=t+v_5(c)$ holds for any $t,k,c \in \mathbb{Z}^+$?
I think that we can prove this by induction on $c$ (by invoking Kummer's theorem at the right time, maybe?), since we have something as: ``$\forall c \in \mathbb{Z}^+, v_5((10^{k+t}+10^t+1)^c-1)=t+v_5(c)$, where $t,k \in \mathbb{Z}^+$''.
Let us start with the base case, so $c=1$.
Now, we know that the $p$-adic valuation is a valuation and we have that $c=1 \Rightarrow v_5((10^{k+t}+10^t+1)^c-1)=v_5(10^{k+t}+10^t)=v_5(10^t \cdot (10^k+1))=v_5(10^t) + v_5(10^k+1)$. Since $10^k+1 \equiv 1 \pmod 5$ for any positive integer $k$, it follows that $v_5(10^k+1)=0$.
Hence, $v_5(10^t) + v_5(10^k+1)=v_5(10^t)=v_5(2^t \cdot 5^t)=v_5(5^t)=t$.
Thus, $c=1 \Rightarrow v_5((10^{k+t}+10^t+1)^c-1)=v_5((10^t+1)^c-1)=t$, $\forall k \in \mathbb{Z}^+$.
The hard part is to prove the inductive step, by assuming that $c \in \mathbb{Z}^+$ is given and supposing that the original statement holds for $v_5((10^{k+t}+10^t+1)^c-1)=t+v_5(c)$.
Unfortunately, the only approach that I have really thinked about is to work with the length of the rightmost repunit "$0$" of $(10^{k+t}+10^t+1)^c$ and then we simply look at the digit on its left (if it is a $5$, then we need to add $2$ to the initial value of the repunit belonging to $10^{k+t}+10^t+1$, otherwise we just add $1$).
How can we avoid this ugly construction and prove the given statement in a strict mathematical sense (the trinomial expansion and then we invoke the generalized form of Kummer's theorem, a long way to go, by any chance)?
Thanks for your time!
