If $\alpha ,\beta $ are cardinals such that $2\leq \alpha \leq \beta$
Prove : $ \beta^{\beta^\alpha} \leq \alpha^{\alpha^\beta}$
My solution : I thought to prove it by induction on $\beta$:
Base case : $\beta = \aleph_0 \implies \aleph_0^{\aleph_0^\alpha} \leq \alpha^{\alpha^{\aleph_0}} \implies \aleph_0 ^ {\aleph_0} \leq \alpha^{\mathfrak c} \implies \mathfrak c \leq \alpha^{\mathfrak c}.$
Induction step : the statement holds for any given cardinality $\beta.$
Prove the statement holds for $\beta+\aleph_0.$
$ (\beta+\aleph_0)^{{(\beta+\aleph_0)}^{\alpha}} \leq \alpha^{\alpha^{\beta+\aleph_0}} \implies (\beta+\aleph_0)^{{\beta}^{\alpha}} \leq \alpha^{\alpha^{\beta}} \implies \beta^{{\beta}^{\alpha}} \leq\alpha^{\alpha^{\beta}}$
(using $\alpha \leq \beta \space \text{then} \space \alpha +\beta=\beta).$
This statement holds.
I am really not sure i supposed to solve this problem by induction and if my induction solution is correct.
I'd be grateful for your feedback!
What are you doing the induction over? $\beta+\aleph_0=\beta$ for any infinite cardinal anyway. And it's not clear what is the index set. Also, why does $\alpha^{\aleph_0}=\frak c$? What happens if $\alpha$ is so much bigger than $\frak c$? (For example, $2^{2^{2^\frak c}}$ is a cardinal.)
The point here is to simply analyse the exponentiation. For example, since $2\leq\alpha\leq\beta$, then $2^\beta\leq\alpha^\beta\leq\beta^\beta$. But if you know that $\beta^\beta=2^\beta$, then equality ensues. And so we get $\alpha^{2^\beta}$, but because of the same reasoning, this is $2^{2^\beta}$.
Since we are interested in weak inequality, we only need to bound the left-hand side from above. Try replacing that lone $\alpha$ by $\beta$ and see what happens.