Prove that for all natural number $x$ and $n$, $x^n - 1$ is divisible by $x-1$.
So here's my thoughts:
it is true for $n=1$, then I want to prove that it is also true for $n-1$
then I use long division, I get:
$x^n -1 = x(x^{n-1} -1 ) + (x-1)$
so the left side is divisible by $x-1$ by hypothesis, what about the right side?
Answer
So first you can't assume that the left hand side is divisible by $x-1$ but for the right hand side we have that $x-1$ divides $x-1$ and by the induction hypothesis we have that $x-1$ divides $x^{n-1}-1$ so what can you conclude about the left hand side.
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