Does $\mathrm{EXP}^\mathrm{EXP}=\mathrm{EXP}$?
Here is my thought: $\mathrm{EXP}$ machine can ask $2^{O(n)}$ queries to the oracle, and each oracle would itself solve an exponential time problem in a single step. So the total power would be $2^{O(n)}\times2^{O(n)}$ which would still be in EXP.
Your definition of EXP is a bit off (you're thinking of $\textbf{E} = \textbf{DTIME}(2^{O(n)})$ instead of $\textbf{EXP} = \textbf{DTIME}(2^{n^{O(1)}})$), but either way the assertion that "each oracle would itself solve an exponential time problem in a single step" is false.
This is because, given exponential time, the machine can write (say) $2^n$ bits to the oracle's input tape, at which point the oracle is allowed to run exponential-time algorithms as measured relative to the size of the $2^n$ bit input (and therefore doubly-exponential in $n$). Thus, just by padding the input to an exponential size and calling the oracle once, you can solve any problem in 2EXP, violating the time hierarchy.
