$\aleph_a^{cf(\aleph_a)} = \aleph_{a+1}$ for $\aleph_a$ regular assuming GCH
$\aleph_a$ regular, so $cf(\aleph_a) = \aleph_a$.
Assuming GCH $2^{\aleph_a} = \aleph_{a+1}$ holds.
But I'm missing the argument for $\aleph_a^{\aleph_a} = 2^{\aleph_a}$. I guess I'm just blind right now. Can someone help?
On
If we assume $\sf GCH$ then $2^\kappa=\kappa^+$ for every $\kappa$, including singular ones.
We have that if $\kappa$ is regular then, $$2^\kappa\leq\kappa^\kappa\leq(2^\kappa)^\kappa=2^\kappa$$
On the other hand if $\kappa$ is singular then $$\kappa<\kappa^+\leq\kappa^{\operatorname{cf}(\kappa)}\leq\kappa^\kappa=2^\kappa=\kappa^+$$
In either case, if we assume $\sf GCH$ we have that $\kappa^\kappa=\kappa^+$.
It is a general fact of cardinal exponentiation that if $2 \leq \kappa \leq \lambda$ and $\lambda$ is infinite, then $\kappa^\lambda = 2^\lambda$. The proof is quite simple: $$2^\lambda \leq \kappa^\lambda \leq ( 2^\kappa )^\lambda = 2^{\kappa \lambda} = 2^\lambda.$$ As you have noted, in your case you have $\mathrm{cf} ( \aleph_a ) = \aleph_a$, since $\aleph_a$ is regular, and so the above is applicable.