PROOF of $e^x > x^m$, $\forall x>1$

Question

Directly using the definition of exp(x) as an infinite series, prove exp(x) ≥ x^m for all x ≥ 1 and m ∈ N.

I have tried truncating the infinite series and showing that the cut off sum is still greater than x^m, which should prove this, but I have always ended up proving some multiple of exp(x) is greater, instead of exp(x) itself.

Note: I am strictly bound to using the summation of x^n/n! definition of exp(x) as a power series and not the (1+x/n)^n definition.

Answer 1

If $e^x\geq x^m$, then $x\geq mln(x)$, or $m\leq \cfrac{x}{ln(x)}$, and since $\cfrac{x}{ln(x)}$ is not bounded and does not have any horizontal asymptotes, there will eventually be some $x$ such that this inequality is satisfied

Answer 2

With series, assuming $x>1,\ $ we have

$e^x-x^m=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots +(\frac{1}{m!}-1)x^m+\frac{x^{m+1}}{(m+1)!}+\cdots\ge$

$ 1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots -x^m+\frac{x^{m+1}}{(m+1)!}+\cdots\ge $

$-x^m+\frac{x^{m+1}}{(m+1)!}$ and this expression is greater than zero if $x$ is large enough.

edit: as Calum points out, the proof is a one liner:

$e^x > \frac{x^{m+1}}{(m+1)!}\Rightarrow e^x -x^m>\frac{x^{m+1}}{(m+1)!}-x^m>0$ if $x$ is large enough.