[ThE-dArK-lOrD]

For $5$, Let $q$ is smallest prime that not in those sequence, We consider $n$ such that $n\cdot (\Pi_{i=1}^{q-2}{p_i^{i!}}) \equiv_q 1$ (Obviously exist)
For this $n$ we get $q\mid n\cdot (p_1^{1!}p_2^{2!}\cdot ... \cdot p_{q-2}^{(q-2)!}\cdot ... \cdot p_n^{n!})+1$ (Since $p_l^{l!} \equiv_q 1$ when $l\geq q-1$ by Fermat's Litttle)
And since all of prime less than $q$ is in this sequence (We can choose $n$ as big as we want), so $p_n=q$, contradiction