A question about Big-O notation
It is straightforward to show that it cannot be correct. As in $O$ notation, any multiplication can be transferred to a non-constant function in exponentiation. A simple counterexample is mentioned by @martini in the comment. When $f(n) = 2n$ and $g(n) = n$, then $f(n) \in O(g(n))$. But, $e^{2n} \not \in O(e^n)$. Hence, $e^{f(n)} \not \in O(e^{g(n)})$.