calculus - 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.

Answer

HINT: The inequality $$\frac{\lambda^n}{e^\lambda} < \frac{C}{\lambda^2}$$ is equivalent to the inequality $\lambda^{n+2}e^{-\lambda}