Prove that $a^n \gt \frac{(n \ln(a))^m}{m!}$ without using taylor series. Prove that $a^n = e^{n \ln(a)} \gt \dfrac{(n \ln(a))^m}{m!}$ for all $m\in \mathbb{N}$ and $a>1$ without using taylor series/calculus. 
I am trying methods that use binomial theorem but I can't get far. I tried induction on $m$. For $m=1$, we have $e^{n\ln(a)}>n\ln(a)$, which is true since $e^x\geq 1+x>x.$ Suppose the proposition is true for $m=t,$ then we have $e^{n\ln(a)}>\dfrac{{(n\ln(a))^t}}{t!}.$ Then in order to prove $m=t+1$ is true we must could show that $e^{n\ln(a)}>\dfrac{e^{n\ln(a)}n\ln(a)}{t+1}.$ But after this step I am stuck. Any hints/ideas will be much appreciated.
 A: Let prove by induction that 
$$e^x >\frac{x^m}{m!} ~~~, \forall ~~m~~ x>0$$


*

*for $n=0$, we have  $e^x > 1 =\frac{x^)0}{0!}$

*Asumme that $e^x  >\frac{x^m}{m!} ~~~,  x>0$


consider $$(0,\infty)\ni x\mapsto f(x) = e^x  -\frac{x^{m+1}}{(m+1)!}$$
Then by Assumption we have $$  f'(x) = e^x  -\frac{x^{m}}{m!}>0$$
that is $f$ is strictly increasing on $(0,\infty)$ therefore, 
$$ 0=f(0)<f(x) =e^x  -\frac{x^{m+1}}{(m+1)!}$$
hence for all $m$
we have $$e^x  -\frac{x^{m}}{m!}>0$$

In particular $a>1$ then, taking, $x= n\ln a>0$ one gets
  $$a^n = e^{n \ln(a)} \gt \dfrac{(n \ln(a))^m}{m!}$$ 

A: Prove with induction. $P(n): e^x>\dfrac{x^n}{n!}$ then $P(0): e^x>1$ is trivial. With assumption $P(m): e^x>\dfrac{x^m}{m!}$ be true, then 
$$\int_0^x e^x>\int_0^x \dfrac{x^m}{m!}=\dfrac{x^{m+1}}{(m+1)!}$$
shows  $P(m+1)$ is true.
A: We assume two results:


*

*$e^x>1+x$ for $x>0$.

*$(e^y)^n= e^{yn}$ for an real $y$ and positive integer $n$.


and prove:

Theorem: If $x>0$ and $m$ a non-negative integer, then $e^x>\frac{x^m}{m!}.$

Assumption (1) implies this result for $m=0,1.$
Given any $x>0$, and any integer $k>m$,  (1) means $e^{x/k}\geq 1+\frac{x}{k}$ and thus, by $(2)$:
$$e^x = (e^{x/k})^k>\left(1+\frac{x}{k}\right)^k\geq 1+\binom{k}{m}\frac{x^m}{k^m}$$
By the binomial theorem.
Now $$\frac{1}{k^m}\binom{k}{m}=\frac{(1-1/k)(1-2/k)\cdots(1-(m-1)/k)}{m!}>\frac{(1-m/k)^m}{m!}$$
Now you need to show that if you pick $\epsilon>0$ you can find $k$ so that:
$$(1-m/k)^m>1-\epsilon$$
(We will assume $\epsilon<1$.)
Then we want $1-\frac{m}{k}>(1-\epsilon)^{1/m}$ or
$$k>m\left(1-(1-\epsilon)^{1/m}\right)^{-1}$$
So this means:
$$e^x>1+\frac{1-\epsilon}{m!}x^m=\frac{x^m}{m!}+\left(1-\frac{x^m}{m!}\epsilon\right)$$ for any $\epsilon>0$.
But then pick $\epsilon <\min\left(1,\frac{m!}{x^m}\right)$ so we get:
$$e^x>\frac{x^m}{m!}$$

None of these steps used calculus, but (1) and (2) depend on our definition of $e$ (or of $e^x$). 
If you define $e^x=\lim_{n\to\infty} \left(1+\frac xn\right)^n$ then you get (1) and (2) pretty easily, but you'd have to prove that this limit exists.
If you define $e^x$ in terms of the power series expansion, then you get the theorem much more directly.
Other ways to define $e^x$ or $e$ are with integrals (definition of natural logarithm) or derivatives, which are calculus. I guess you could "hide" the calculus by defining natural log as a limit of the Reimann sums, and brute force from there.
