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