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.

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 $\alpha(a^{n+1}-1) < \alpha a^{n+1}$.

