$\mathsf{2EXP} = \mathsf{EXP}^{\mathsf{EXP}}$?
Consider the problem 
$$A^f = 
A_{TM}^f = 
\{\langle M,x,t \rangle \mid 
M\in DTM \text{ and } 
M(x) \text{ halts and accepts in } f(|t|) \text{ steps}\}.$$

Now 
$L^{\exp}$ is complete for $\mathsf{EXP} = \mathsf{DTime}(\exp (n^{O(1)}))$ and
$L^{\exp\exp}$ is complete for $\mathsf{2EXP} =\mathsf{DTime}(\exp\exp(n^{O(1)}))$.

We show that $L^{\exp \exp}$ is in $\mathsf{EXP}^\mathsf{EXP}$.

Given $\langle M,x,t \rangle$, 
we simply write $\langle M,x,1^{\exp(|t|)} \rangle$ on the query tape and 
ask it from $L^{\exp}$ and return its answer as output.

This algorithm is in $\mathsf{EXP}^\mathsf{EXP}$, therefore $\mathsf{2EXP} \subseteq \mathsf{EXP}^\mathsf{EXP}$.