@The-Blade-Dancer The factor of \(2^{19}\) comes from all the even numbers in the factorial \(N!,\) like \(2, 4, 6, 8, 12, 14, \ldots\) etc., in addition to extra \(2\)'s that come from the powers of \(2\) like \(4, 8, 16.\)
Actually, though, you know that \(N\) must equal \(22,\) since we have two \(11\)'s (from the fact that \(11^2\) is a factor). One is from \(11,\) and the other is from \(22.\) We know that \(N\) can't be larger than \(22,\) so \(\boxed{N=22}.\)
And then \(N!\) (N factorial) would be \(22! = 1 \times \textcolor{red}{2} \times 3 \times \textcolor{red}{4} \times 5 \times \textcolor{red}{6} \times \textcolor{blue}{7} \times \textcolor{red}{8} \times 9 \times \textcolor{red}{10} \times \textcolor{yellow}{11} \times \textcolor{red}{12} \times 13 \times \textcolor{purple}{14} \times 15 \times \textcolor{red}{16} \times 17 \times \textcolor{red}{18} \times 19 \times \textcolor{red}{20} \times \textcolor{blue}{21} \times \textcolor{orange}{22}.\)
I tried to color-code so that all of the \(\textcolor{red}{\text{red}}\) numbers are even and contribute factors of \(2\) (this is where you get the \(2^{19}\)), the \(\textcolor{blue}{\text{blue}}\) numbers contribute factors of \(7\), and the yellow numbers contribute factors of \(11.\) Since \(\textcolor{purple}{14}\) is a multiple of both \(2\) and \(7,\) I made it purple. And since \(22\) is a multiple of both \(11\) and \(2,\) I made it orange.