If $d, e$ are infinite cardinals, then does $2^d \geq 2^e$ imply $d \geq e$?

Question

For infinite cardinals $d$ and $e$, does $2^d \geq 2^e$ imply that $d \geq e$?

I would like a proof or a counterexample. Background so far: The cardinals form a chain under $\leq$; if $e$ is an infinite cardinal with $d \leq e$, then $de = e$ and $d + e = e$; for any cardinals we have $d^{e_1 + e_2} = d^{e_1} d^{e_2}$, $(d_1d_2)^e = d_1^e d_2^e, (d^e)^f = d^{ef}$.

Thank you in advance for any hints or answers.

Answer 1

If you do not assume GCH, then it can happen that $2^{\aleph_0} = 2^{\aleph_1} = \aleph_2$, which gives you a counter-example $d = \aleph_0$ and $e = \aleph_1$.