We will prove by mathematical induction that for all positive integers $n$, $5^n - 1$ is divisible by 4. In other words, we want to show that for every positive integer $n$, there exists an integer $k$ such that:

$5^n - 1 = 4k$

Step 1: Base Case

We begin with the base case $n = 1$.

$5^1 - 1 = 5 - 1 = 4$

Since 4 is divisible by 4, the base case holds.

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Step 2: Inductive Hypothesis

Assume that the statement is true for some positive integer $n = k$. That is, assume:

$5^k - 1 = 4m$

for some integer $m$. This is our inductive hypothesis.

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Step 3: Inductive Step

We need to show that the statement holds for $n = k + 1$. That is, we need to prove:

$5^{k+1} - 1 \text{ is divisible by 4}.$

We start by manipulating $5^{k+1} - 1$ using the properties of exponents:

$5^{k+1} - 1 = 5 \cdot 5^k - 1.$

Now, add and subtract 5 in the expression:

$5 \cdot 5^k - 1 = (5 \cdot 5^k - 5) + 4.$

Factor out 5 from the first part:

$(5(5^k - 1)) + 4.$

By the inductive hypothesis, we know that $5^k - 1 = 4m$ for some integer $m$. Substituting this into the expression:

$5(4m) + 4 = 20m + 4 = 4(5m + 1).$

Since $5m + 1$ is an integer, the expression $4(5m + 1)$ is divisible by 4.

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Step 4: Conclusion

By mathematical induction, we have shown that $5^n - 1$ is divisible by 4 for all positive integers $n$.

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Do you have any questions or need further clarifications?

Here are 5 related questions for further exploration:

1. How would the proof change if the statement involved divisibility by 8 instead of 4?
2. Can you prove that $5^n + 1$ is divisible by 3 for odd $n$?
3. How is mathematical induction applied differently in strong induction?
4. What is the relationship between divisibility proofs and modular arithmetic?
5. Can we generalize the result for powers of other numbers, like $6^n - 1$?

Tip: When using mathematical induction, always carefully link the inductive step to the hypothesis to avoid logical gaps.