Show that for all $\lambda \geq 1~$ $\frac{\lambda^n}{e^\lambda} < \frac{C}{\lambda^2}$ 
Show that for any $n \in \mathbb N$ there exists $C_n > 0$ such that for all $\lambda \geq 1$
$$ \frac{\lambda^n}{e^\lambda} < \frac{C_n}{\lambda^2}$$

I can see that both sides of the inequality have a limit of $0$ as $\lambda \rightarrow \infty$ since, on the LHS, repeated application of L'Hôpital's rule will render the $\lambda^n$ term as a constant eventually, while the $e^{\lambda}$ term will remain, and the RHS is obvious.
I can also see that the denominator of the LHS will become large faster than the RHS denominator, but I can't seem to think of anything that will show that the inequality is true for all the smaller intermediate values.

The function $\lambda \mapsto \lambda^{n+2}$ is strictly increasing for positive $\lambda$ and also $e^{\lambda} > \lambda$. Combining this you get
$$e^{\lambda} = \left( e^{\frac{\lambda}{n+2}} \right)^{n+2} > \left( \frac{\lambda}{n + 2} \right)^{n+2}$$
for all positive $\lambda$ and therefore
$$\frac{\lambda^n}{e^\lambda} < \frac{\lambda^n (n+2)^{n+2}}{\lambda^{n+2}} = \frac{(n+2)^{n+2}}{\lambda^2}.$$
