I hope this is not counted as a duplicate, as I would like to know if my proof is valid:
$P(n): 5^n - 1$ is divisible by $4$ for $n \ge 0$.
Base Step: $P(0): 5^0-1 = 1-1 = 0 = 0\times 4$.
Induction Supposition: $P(k): 5^k-1$ is divisible by $4$.
Prove: $P(k+1): 5^{k+1}-1$ is divisible by $4,$ or equivalently $5^{k+1}-1 = 4r$, for some integer $r$.
$5^{k+1}-1$
$= 5^k\times5-1$ by Exponent Laws
$= 5\times4r$ by I.H.
$=4(5r)$ which was to be shown.
...or without direct induction, using
$$A^n-B^n=(A-B)\left((A^{n-1}+A^{n-2}B+\ldots+AB^{n-2}+B^{n-1}\right)$$
...which is proved, of course, with induction... :) , and then
$$5^n-1=5^n-1^n=\overbrace{(5-1)}^{=4}(5^{n-1}+5^{n-2}+\ldots+5+1)$$
which is then trivially seen to be divisible by $\;4\;$ .
The way you did it was pretty close too, but your line before the last one is wrong:
$$5^{k+1}-1=5\cdot5^k-1=4\cdot5^k+(5^k-1)$$
and first summand is trivially divisible by four, whereas the second one is by the inductive hypotheses.
It'd be better if you showed the transition between your "exponent law" step and "I.H." step clearly as follows:
$$5^{k+1}-1=5\times 5^k-1=5\times (4p+1)-1=4(5p+1)\equiv 0\pmod4$$
where $p$ is some integer taken in the I.H. to represent divisibility by $4$.
Just demonstrating an alternate solution, this can be proved in one line using modular arithmetic.
$$\forall~n\in\Bbb{Z_0^+}~,~5^{n}-1=(4+1)^n-1\equiv 1^n-1\equiv 1-1\equiv 0\pmod4$$
Your proof is wrong because
$$5^k\cdot5-1\ne5\cdot4\cdot r$$ (take $k=1$ or more as a counterexample.)
It seems that you confused $$5^k\cdot5-1$$ with $$5\cdot(5^k-1),$$ which is quite different.
The correct deduction is $$5^k\cdot5-1=(5^k-1+1)\cdot5-1=(4r+1)\cdot5-1=4\cdot5\cdot r+4=4.(5\cdot r+1)=4\cdot r'.$$
For $n=1$ we have $5^{n} - 1 = 5 - 1 = 4.$
Suppose there is an $n \geq 1$ such that $5^{n} - 1 = 4k$ for some $k$. Then $$5^{n+1} - 1 = 5(5^{n} - 1) + 4 = 20k + 4,$$ so $$4 \mid (5^{n+1} - 1).$$
Base case is OK. Now assume that for any $n=k$, $5^k-1=4p$ --(i) for an integer p.
We wish to prove that $5^k.5-1=4q$ for an integer q.
$5^k.5-5+5-1=4q$
$5(5^k-1)+4=4q$
$5.4p+4=4q$ (by (i))
$4(5p+1)=4q$
$q$ is an integer. So $5p+1=q$ will also be an integer. Hence proved.
Your application of the induction hypothesis is wrong: you know that $5^k-1=4r$, and you are basically doing $$ ab-1=(a-1)b $$ which you clearly see is incorrect.
But you're almost there! You can write $5^k=4r+1$, so $$ 5^{k+1}-1=5\cdot 5^k-1=5(4r+1)-1=20r+5-1=20r+4=4(5r+1) $$
Comment: As others have mentioned, where you went wrong was going from $5^k\times 5-1$ to $5\times 4r$. If I were tasked with proving that $5^n-1$ is divisible by $4$ for all $n\geq 0$, then I would use Timbuc's method for certain because it minimizes legwork, but I imagine you were asked to prove the problem in question using an inductive argument. As written about before, writing up an induction proof is a bit of an art form. The more advanced you become and the more advanced your readership is, the more you can afford to leave out many "unimportant" details such as "trivial derivations" and the like, but I doubt that is what is expected of you here. On the contrary, I imagine your goal is to provide a clear, clean induction proof, and that is really the only reason why I am answering--I thought I would try to provide a clear, polished way of proving your result using induction. The proof outlined below is virtually identical to the proof given here where the goal was to prove that $6^n-1$ is divisible by $5$ for all $n\geq 1$. Hopefully you will find what follows below to be helpful and, most of all, clear.
For $n\geq 0$, let $S(n)$ denote the statement $$ S(n) : 4\mid(5^n-1)\Longleftrightarrow 5^n-1=4m, m\in\mathbb{Z}. $$ Base case ($n=0$): $S(0)$ says that $4\mid(5^0-1)$, and this is true.
Inductive step: Fix some $k\geq 0$ and assume that $S(k)$ is true where $$ S(k) : 4\mid(5^k-1)\Longleftrightarrow 5^k-1=4\ell, \ell\in\mathbb{Z}. $$ To be proved is that $S(k+1)$ follows where $$ S(k+1) : 4\mid(5^{k+1}-1)\Longleftrightarrow 5^{k+1}-1=4\eta, \eta\in\mathbb{Z}. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 5^{k+1} - 1 &= 5^k\cdot 5-1\tag{by definition}\\[0.5em] &= (4\ell+1)\cdot 5-1\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 5\cdot 4\ell+4\tag{expand}\\[0.5em] &= 4(5\ell+1)\tag{factor out $4$}\\[0.5em] &= 4\eta.\tag{$\eta=5\ell+1; \eta\in\mathbb{Z}$} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 0$. $\blacksquare$
