Some Mathematical Induction

Here is a problem similar to one was posed in a YouTube video that I watched recently:

Prove that $a^n-b^n$ is divisible by $a-b$ for all positive integer values of $n$ with $a$ and $b$ both positive integers and $a \gt b$.

We can do this using mathematical induction as follows:

It's clearly true when $n=1$ and so let's assume it's true for some value of $k \gt 1$. This means that:

$$\frac{a^k-b^k}{a-b}=m$$ where $m$ is an integer. Can we now show that it's true for $k+1$. Let's begin: $$\begin{align} \frac{a^{k+1}-b^{k+1}}{a-b} &= \frac{a \, . b^k-b \, . b^k}{a-b} \\ a-b &= c \text{ where c is some positive integer} \\ a & = b+c \\ \frac{a \, . b^{k}-b \, . b^{k}}{a-b} &= \frac{(b+c) \, . b^{k}-b \, . b^{k}}{c} \\ &= \frac{b \, . b^{k} +c \,.  b^{k} - b \, . b^{k}}{c} \\ &= \frac{c \, . b^{k} }{c} \\ &=b^{k} \text{ which is an integer} \end{align}$$ Since it is true for $n=1$ and $n=k+1$ when it is true for $n=k$ then it is true for all $n$.

In the video that I referred to at the start of this post, a specific instance of the more general situation was looked at, namely:

$$\text{prove that } \frac{11^n-4^n}{7} \text{ is always an integer}$$