# 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

• proofwiki.org/wiki/Sum_of_Geometric_Progression – lab bhattacharjee Aug 17 '14 at 9:53
• You don't need induction. Just repeat the proof as in the case $n=2$. – Quang Hoang Aug 17 '14 at 9:53
• Prove it for n=1; then, assuming that it is true for n=k, try to show that it is true for n=k+1. It is easy indeed. (I do not know the name of this proving method.) – Enthusiastic Engineer Aug 17 '14 at 10:11
• @EnthusiasticEngineer It is called Proof by Induction – BadAtAlgebra Feb 17 '19 at 7:02

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 !

• Your sums should end in $\dots+x^2+x$, not $\dots+x+x$. – Joonas Ilmavirta Aug 17 '14 at 10:10
• Ahh yes ! will fix it thanks :) – AgentS Aug 17 '14 at 10:11

Use the formula for sum of a geometric series: $$1+x+\ldots +x^{n-2}+x^{n-1}=\frac{x^n-1}{x-1}$$

• Isn't geometric series formula derived from what is to be proven here? – zbrojny120 Apr 20 at 15:14

You may just open hooks in right half of expression: $$(x - 1)(x^{n - 1} + x^{n - 2} + \dots + x + 1) = x * (x^{n - 1} + x^{n - 2} + \dots + x + 1) - 1 * (x^{n - 1} + x^{n - 2} + \dots + x + 1) = (x^n + x^{n - 1} + \dots + x^2 + x) - (x^{n - 1} + x^{n - 2} + \dots + x + 1) = x^ n - 1$$ No problems)

$$(x−1)(x_n−1+x_n−2+...+x+1)=x^n+x^{n-1}+.....+x^2+x-[x^{n-1}+x^{n-2}+....+x+1]=x^n-1$$