Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$

So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$

I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $(x-1)(x+1)$ which when distributed is $x^2-1$. So my base case holds.

Now I assume that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ for some $n$. However, this is where I am stuck. Am I trying to show $x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)$? I am still a novice when it comes to these induction proofs. Thanks

Solution 1:

To conclude your induction proof, just multiply x both sides :

$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1) $

multiply $x$ both sides :

$\begin{align} \\ x^{n+1}-x &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 -(x-1) &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x)+(x-1) \\ \end{align}$

factor $(x-1)$ and you're done !