Prove $x^n\ge 1+n(x-1) \quad \forall x \ge 1, \forall n \in N$ My attemp:
using the induction principle :
if $n=1$ I have an identity $x\ge x$.I suppose the principle is true for $n$ and I want to prove it for $n+1$
$x^{n+1} = x^n \cdot x\ge [1+n(x-1)]\cdot x$ but how can I obtained $1+(n+1)(x-1)$?

Here is a one liner,
$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\geq (x-1) \underbrace{(1+\cdots +1)}_\text{n of them}=n(x-1)$
