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.

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
$$\begin{align*}
\frac{k^k}{k!} 
& = \left(\frac{k}{k-1}\right)^{k-1}
\left(\frac{k-1}{k-2}\right)^{k-2}
\left(\frac{k-2}{k-3}\right)^{k-3}\cdots
\left(\frac{4}{3}\right)^3
\left(\frac{3}{2}\right)^2
\left(\frac{2}{1}\right)\\
& < e \cdot e \cdot e \cdots \cdots e \cdot e \cdot e
= e^{k-1} < e^k \end{align*}
$$
