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.
 A: HINT: The inequality $$\frac{\lambda^n}{e^\lambda} < \frac{C}{\lambda^2}$$ is equivalent to the inequality $\lambda^{n+2}e^{-\lambda}<C$. Consider the function $f(x)=x^{n+2}e^{-x}$ on the positive real line; does it have a maximum somewhere?
A: Considering the power series of exp seems to give the simplest solution. For $\lambda>0$ we have
$$
e^\lambda = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!}
$$
$$
e^\lambda > \frac{\lambda^{n+2}}{(n+2)!}
$$
$$
\frac{(n+2)!}{\lambda^2} > \frac{\lambda^{n}}{e^\lambda}
$$
so that $C=(n+2)!$. As Brian M. Scott remarks in a comment above, $C$ necessarily depends on $n$ as $\lambda^n/e^\lambda\rightarrow\infty$ as $n\rightarrow \infty$ for $\lambda > 1$.
A: 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
$$\frac{\lambda^n}{e^\lambda} < \frac{\lambda^n (n+2)^{n+2}}{\lambda^{n+2}} = \frac{(n+2)^{n+2}}{\lambda^2}.$$
