showing an inequality not using stirling formula I don't know how to show that
$$
\frac{k^k}{k!}\leq e^{k}
$$
without using Stirling's approximation. I want to show it directly. I guess I need some inequality to achieve this but I don't know.
 A: Both sides of the inequality are positive. So log both sides, and prove $k\ln k-\ln(k!)\le k$ :
$$\ln(k!)=\sum_{i=2}^k\ln i\ge\sum_{i=2}^k\int_{i-1}^i\ln x\,dx=\int_1^k\ln x\,dx=\left[x\ln x - x\right]_1^k =(k\ln k-k)-(-1)\ge k\ln k-k
$$
A: And yet another approach. First note that for $n>k$
$$ \prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)=\frac{n}{n}\frac{n-1}{n}\ldots\frac{n-k+1}{n}=\frac{n!}{(n-k)!n^k}\rightarrow 1\tag{1}$$
when $n\rightarrow\infty$ because $k$ is fixed and each term of the product converges to 1.
Therefore, since you know that (by the binomial expansion)
\begin{align} \left(1+\frac{k}{n}\right)^n&=1+\binom{n}{1}\frac{k}{n}+\ldots+\binom{n}{k}\left(\frac{k}{n}\right)^k+\ldots+\binom{n}{n}\left(\frac{k}{n}\right)^n\\
&\ge \binom{n}{k}\left(\frac{k}{n}\right)^k \\
&=\frac{n!}{(n-k)!n^k} \left(\frac{k^k}{k!}\right)\end{align}
Taking limits on both sides you have (depends on what definition you have of $e^k$ but usually this is the most elementary)
$$ e^k=\lim_{n\rightarrow\infty} \left(1+\frac{k}{n}\right)^n\ge\left(\frac{k^k}{k!}\right)\lim_{n\rightarrow\infty}\frac{n!}{(n-k)!n^k}=\left(\frac{k^k}{k!}\right)~~,$$
because of (1).
A: Suppose
$\frac{k^k}{k!}\leq e^{k}
$
and
$\frac{(k+1)^{k+1}}{(k+1)!}> e^{k+1}
$.
Then
$e
=\frac{e^{k+1}}{e^{k}}
<\frac{\frac{(k+1)^{k+1}}{(k+1)!}}{\frac{k^k}{k!}}
=(1+\frac1{k})^k
$.
But
$(1+\frac1{k})^k
< e
$,
as shown in many places
(including this question of mine:
What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?).
Therefore,
$\frac{k^k}{k!}\leq e^{k}
\implies
\frac{(k+1)^{k+1}}{(k+1)!}> e^{k+1}
$.
Since
$\frac{1^1}{1!}
< e^1
$,
$\frac{k^k}{k!}\leq e^{k}
$ for all $k$.
A: A somewhat different approach:
In the pmf for a Poisson$(k)$ $P(X=k) = \frac{e^{-k} k^k}{k!}$ but $P(X=k)<1$ (since it's a probability).
The result follows by multiplying by $e^k$.
A: The Taylor series for $e^x$ is $$e^x = \sum\limits_{n=0}^\infty \frac{x^n}{n!}$$
so for an integer $k$ we have 
$$e^k = \sum\limits_{n=0}^\infty \frac{k^n}{n!} = 1+\frac{k}{1!}+\frac{k^2}{2!}+\cdots+\frac{k^k}{k!}+\cdots$$
If you don't want to use Taylor series, you could prove it with induction. All we'll need is the Binomial Theorem, some algebra skills, and a small (easy to verify) fact at the end.
Step one (starting with $k=1$):
$$e \geq 1$$
Great, let's proceed to the induction step, where we suppose that $e^k \geq \frac{k^k}{k!}$ for some (fixed) $k$ no less than $2$:
\begin{align}\frac{(k+1)^{k+1}}{(k+1)!} &= \frac{(k+1)^k}{k!} \\ &= \frac{\sum_{i=0}^k \binom{k}{i}k^{k-i}}{k!} \\ &= \sum_{i=0}^k \frac{1}{k!}\binom{k}{i}k^{k-i} \\ &= \sum_{i=0}^k \frac{k^{k-i}}{i!(k-i)!} \\
&= \frac{k^k}{k!}+\frac{k^{k-1}}{1!(k-1)!}+\frac{k^{k-2}}{2!(k-2)!}+\cdots+\frac{1}{k!} \\ &\leq e^k+\frac{e^{k-1}}{1!}+\frac{e^{k-2}}{2!}+\cdots+\frac{1}{k!} \\ &< e^k + e^{k-1} + \cdots + 1\end{align}
The last sum can be shown to be less than $e^{k+1}$ provided you can show that, for all real numbers $a \geq 2$, the inequality $a^{n+1} > a^n+a^{n-1}+\cdots+a+1$ holds for all $n \in \mathbb{N}$. 
A short proof of that last claim, for those interested:

 One can easily show that $a^n+a^{n-1}+\cdots+a+1 = \frac{a^{n+1}-1}{a-1}$. Let $\alpha = \frac{1}{a-1}$. Then $a \geq 2 \Rightarrow 0 < \alpha \leq 1 \Rightarrow \alpha a^{n+1} \leq a^{n+1} \Rightarrow \frac{a^{n+1}-1}{a-1} = \alpha(a^{n+1}-1) < a^{n+1}$ since $\alpha(a^{n+1}-1) < \alpha a^{n+1}$.

A: If you want a somewhat elementary argument not using Taylor series, I have one that only relies on the fact that for any $x>0$ we have $\left(1 + \frac{1}{x}\right)^x < e$ (which is proved by showing $\ln(x+1)-\ln(x) < \frac{1}{x}$ using integration).
Then we have
$$\begin{align*}
\frac{k^k}{k!} 
& = \left(\frac{k}{k-1}\right)^{k-1}
\left(\frac{k-1}{k-2}\right)^{k-2}
\left(\frac{k-2}{k-3}\right)^{k-3}\cdots
\left(\frac{4}{3}\right)^3
\left(\frac{3}{2}\right)^2
\left(\frac{2}{1}\right)\\
& < e \cdot e \cdot e \cdots \cdots e \cdot e \cdot e
= e^{k-1} < e^k \end{align*}
$$
A: Since
$e^x
=\sum_{n=0}^{\infty} \frac{x^n}{n!}
$,
for $x > 0$
and any integer $n$,
$e^x
> \frac{x^n}{n!}
$.
Setting $x = n$,
$e^n
> \frac{n^n}{n!}
$.
