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 $a>b$.

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 $k>1$. 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:$$\begin{array}{rl}\frac{a^{k+1}-b^{k+1}}{a-b}& =\frac{a.b^k-b.b^k}{a-b}\\ a-b& =c\text{ where c is some positive integer}\\ a& =b+c\\ \frac{a.b^k-b.b^k}{a-b}& =\frac{(b+c).b^k-b.b^k}{c}\\ & =\frac{b.b^k+c.b^k-b.b^k}{c}\\ & =\frac{c.b^k}{c}\\ & =b^k\text{ which is an integer}\end{array}$$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:$$\text{prove that }\frac{{11}^n-4^n}{7}\text{ is always an integer}$$