# Math Help - Proof on Powers of Numbers Greater Than 1

1. ## Proof on Powers of Numbers Greater Than 1

I'm working on a proof and I'm stuck at what seems to me a severe case of mathematician's block (which is sort of like writer's block). I need to prove that when you take a number greater than 1 and raise it to any power the result is going to be larger than the original number. In other words:

If $x \ge 1$ and $n$ is a natural number, then $x^n \ge x$.

2. Originally Posted by eikou
If $x > 1$ and $n$ is a natural number, then $x^n > x$.
Well of course it should be $x^n\ge x$. What if $n=1?$

Is this true $x^{n-1}\ge 1?$

3. Right right, greater than or equal to. With the natural numbers, there will be no $x^{n-1}$ though (beneath 1), because they start at 1 and count upwards by integers. A condition on the proof was that the exponent belong to the natural numbers. Therefore, by consequence, $x^{n-1} \ge 1$ will always hold true since $n-1 \ge 1$. Otherwise the conditions of the proof are violated and it's... what's that called... vacuously true I think.

4. Let $f:[1,\infty)\mapsto\mathbb{R}$ be given by $f(x)=x^n-n,\text{ }n\in\mathbb{N}$. We have that $f(1)=0$ and $f'(x)=nx^{n-1}-1\geqslant n-1\geqslant 1-1=0$. Where it follows that $f(x)\geqslant f(0)=0$ for all $x\in[1,\infty)$

5. Alternatively, $x^n-x==x\left(x^{n-1}-1\right)\geqslant x\left(1-1\right)=0$.

6. You could also do induction. The possibilities are endless.

7. I actually tried induction first, epic flop. Not to say you can't, just that it was beyond my ability. I tend to run into difficulty frequently with induction proofs. I always get stuck in vicious circles.

8. Originally Posted by eikou
I actually tried induction first, epic flop. Not to say you can't, just that it was beyond my ability. I tend to run into difficulty frequently with induction proofs. I always get stuck in vicious circles.
Really? Seriously? Are you sure that you didn't make a careless clerical error?

Assume that $x^n\geqslant x$ for all $x\geqslant1$ then $]x^{n+1}\geqslant x^2\geqslant x$. Induction. Over.