Is $2^{2n} = O(2^n)$? Is $2^{2n} = O(2^n)$?
My solution is:
$2^n 2^n \leq C_{1}2^n$
$2^n \leq C_{1}$,
TRUE.
Is this correct?

If $2^{2n}=O(2^n)$, then there is a constant $C$ and an integer $M$ such that for  all $n\ge M$, the inequality $2^{2n}\le C 2^n$ holds.
This would imply that  $2^n\cdot 2^n\le C 2^n$ for all $n\ge M$, which in  turn implies $$\tag{1} 2^n\le C  \quad  {\bf for\ all } \quad n\ge M. $$ Can such $C$ and $M$ exist? Note the right hand side of $(1)$ is fixed, and the left hand side...
