calculus - How to show $lim_{n to infty}a_n=frac{5^{3cdot n}}{2^{left(n+1right)^2}}$?

$$\lim_{n \to \infty}a_n=\dfrac{5^{3\cdot n}}{2^{\left(n+1\right)^2}}$$

I am trying to solve it using the squeeze theorem.
I have opened the expression to $$a_n=\dfrac{5^3\cdot 5^n}{2^{n^2}\cdot2^{2n}\cdot2)}$$
I think that the LHS should be $$a_n=\dfrac{2^3\cdot 2^n}{2^{n^2}\cdot2^{2n}\cdot2)}$$
But as for the RHS I do not find a bigger expression, any ideas?

Answer

Hint: $0<\dfrac{125^n}{2^{(n+1)^2}} < \dfrac{125^n}{2^{n^2}} < \left(\dfrac{1}{2}\right)^n$

since $\dfrac{125}{2^n} < \dfrac{1}{2}$ when $n > 8$.