What is the limit $\lim\limits_{n \to \infty} \frac{n^{\sqrt{n}}}{2^n}$? Question:

Prove that
$\displaystyle
\lim\limits_{n \to \infty} \frac{n^{\sqrt{n}}}{2^n} = 0.
$

I was thinking of using the Squeeze Theorem (might not be the right way to go), but finding an upper-bound function proved to be quite tricky.
 A: For a 'more' elementary proof (which does not involve $e$ or $\log$), 
Hint:
Prove using induction that for any $k \gt 7$, we have that
$$2^{k-1} \gt (k+1)^2$$
Try using this with your squeeze theorem idea.
For completeness:
Let $k = \lfloor \sqrt{n} \rfloor$. Using the above we have that, for all $n \gt 100$,
$$2^{\sqrt{n} - 1} \gt 2^{k-1} \gt (k+1)^2 \gt n$$
Raising to $\sqrt{n}^{th}$ power we get
$$2^{n - \sqrt{n}} \gt n^{\sqrt{n}}$$
and so
$$ 2^{-\sqrt{n}} \gt \frac{n^{\sqrt{n}}}{2^n}$$
A: Thanks Zev and Dan, I was on this trail. My answer goes like this:
I compare the rate of change of both functions (1) $\ln(x){ \sqrt{x} }$ and (2) ${\ln(2)}x$ by taking their derivatives and evaluating them as x goes to infinity : 
(1)
$$
\lim_{x \to \infty} \frac{1}{\sqrt x} + {1/2}\frac{\ln {x}}{\sqrt x} = 0
$$
Tip: use L'Hôpital's rule if you were unsure about the 2nd quotient.
(2)
$$
\lim_{x \to \infty} \ln {2} = \ln {2}
$$
So (2) has a greater growth than (1) for x going to infinity so
$$
\lim_{x \to \infty}\frac{x^{\sqrt{x}}}{2^x}=\frac{e^{\ln(x)\sqrt{x}}}{e^{\ln(2)x}}=e^{\ln(x){ \sqrt{x} }-\ln(2)x} = 0
$$
A: Let's take n= m^2 and get $\lim\limits_{m \to \infty} \frac{m^{2m}}{2^{m^2}}$. By ratio test we have that:
$$R=\lim\limits_{m \to \infty} \frac{(m+1)^{2m+2}}{2^{(m+1)^2}}\cdot\frac{2^{m^2}}{m^{2m}} =\lim\limits_{m \to \infty}\frac{e^2 (m+1)^2}{2^{2m+1}}=0.$$
Q.E.D.
A: HINT $\rm\displaystyle\quad \dfrac{X^{\sqrt X}}{2^X}\ =\ \bigg(\dfrac{Z^2}{{\it e}^{\:c\:Z}}\bigg)^{\!\!Z} =:\ F(Z)^Z,\quad \begin{array}* Z\: =\: \sqrt{X}\\ \rm c\: =\: \ln\:2\end{array}\ \ $ and $\displaystyle\rm\ \lim_{Z\:\to\:\infty}\ F(Z)\: = 0\ $ by l'Hopital.
