Question
Show that for $n>0$,
$n.2^{2^{\aleph _0}}=\aleph _0.2^{2^{\aleph _0}}= 2^{\aleph _0}.2^{2^{\aleph _0}}= 2^{2^{\aleph _0}}.2^{2^{\aleph _0}}=(2^{2^{\aleph _0}})^n=(2^{2^{\aleph _0}})^{\aleph _0}= (2^{2^{\aleph _0}})^{2^{\aleph _0}}=2^{2^{\aleph _0}}.$
Attempt
$2^{2^{\aleph_ 0}}\leq n.2^{2^{\aleph _0}}\leq {\aleph _0}.2^{2^{\aleph _0}}\leq 2^{\aleph _0}.2^{2^{\aleph _0}}\leq 2^{2^{\aleph _0}}.2^{2^{\aleph _0}}=2^{2^{\aleph _0}+2^{\aleph _0}}=(2^{2.2^{\aleph _0}})=(2^{2^1.2^{\aleph _0}})=2^{2^{1+\aleph _0}}=2^{2^{\aleph_0}}.$
I feel know this can be done with about three sequences of inequalities. So, I am not sure of the fourth inequality. Also, I think some things may be missing, as I am not completely confident of this. Your help would be appreciated.