Prove by induction that $5^n - 1$ is divisible by $4$.

We prove that for all $n \in \mathbb{N}$, $4 \mid \left( 5^n-1 \right)$. (Notationally, this says $4$ divides $5^n-1$ with a zero remainder).

1. For a basis, let $n=1$. Then $$5^1-1=4,$$ and clearly $4\mid4$.

2. Assume that $5^n-1$ is is divisible by $4$ for $n=k, \, k \in\mathbb{N}$. Then by this assumption, $$4 \mid \left( 5^k-1 \right) \Rightarrow 5^k-1=4m, \, m \in \mathbb{Z}.$$ (This notationally means that $5^k-1$ is an integer multiple of $4$.)

3. Let $n=k+1$. Then $$ \begin{align*} 5^{k+1}-1 &= 5^k \cdot 5-1 \\ &=5^k(4+1)-1 \\ &=4\cdot 5^k+5^k -1 \\ &=4\cdot5^k+4m\\ &=4\left( 5^k+m \right). \end{align*} $$ Since $4\mid4\left( 5^k+m \right)$, we may conclude, by the axiom of induction, that the property holds for all $n \in \mathbb{N}$.

First prove the base case $n=1$. Then induct and make use of the fact that $$(5^{n+1}-1) - (5^n-1) = 4 \cdot 5^n$$to conclude what you want.

Without induction, you can use the identity

$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)$$

Of course you would still need induction or something to prove this identity.

without induction $$5^n-1=(4+1)^n-1$$ $$=4^n+n4^{n-2}+...+1-1$$ the only term in $(1+4)^n$ not being multiplied by a power of $4$ is $1$ but it disappears due to the $-1$.

Why induction? $5^n$ ends in $\dots25$ for $n>1$, so $5^n-1$ ends in $\dots24$.