If $2^{\aleph_{\beta}}\geq \aleph_{\alpha}$, then $\aleph_{\alpha}^{\aleph_{\beta}}=2^{\aleph_{\beta}}$. If $2^{\aleph_{\beta}}\geq \aleph_{\alpha}$, then $\aleph_{\alpha}^{\aleph_{\beta}}=2^{\aleph_{\beta}}$. 
Proof: Note that if $\beta \geq \alpha$, then we have $\aleph_{\alpha}^{\aleph_{\beta}}=2^{\aleph_{\beta}}$. I don't know whether the condition given $2^{\aleph_{\beta}}\geq \aleph_{\alpha}$ will give $\beta \geq \alpha$ or not. It it does, then we are done. If it is not, then we still have to consider other cases.
UPDATE: Clearly $2 ^ {\aleph_{\beta}} \leq \aleph_{\alpha}^{\aleph_{\beta}}$. Note that $\aleph_{\alpha}^{\aleph_{\beta}} \leq (2^{\aleph_{\beta}})^{\aleph_{\beta}}=2^{\aleph_{\beta}}$. By Cantor-Bernstein Theorem, we have the desired result. 
Can anyone help me to check whether my update is correct or not.

You don't necessarily have that $2^{\aleph_\beta}\geq\aleph_\alpha$ implies $\beta\geq\alpha$; consider that $2^{\aleph_0}\geq\aleph_1$, but $0$ certainly isn't $\geq 1$.  Instead, try proving that you have both $\aleph_\alpha^{\aleph_\beta}\geq 2^{\aleph_\beta}$ and $\aleph_\alpha^{\aleph_\beta}\leq2^{\aleph_\beta}$ and then using Cantor-Bernstein.  One of these should be trivial; for the other you'll have to do a little cardinal arithmetic using your hypothesis.
