Number Theory : Infinitude Of Primes - a different proof

The proof implies that given any prime $\,\color{#c00}p\,$ there exists a larger prime (dividing $\,Q_{\large\color{#c00}p}$), therefore the set of primes is infinite.

Remark $\ $ Because this way of proof is not by contradiction, it yields constructive information: iterating the above yields an algorithm to generate an infinite sequence of primes, viz.

$$\begin{align} &Q_1 = 1!+1 = 2\quad\ \text{has prime factor}\ \ \ \, 2 > 1\\ &Q_2 = 2!+1 = 3\quad\ \text{has prime factor}\ \ \ \, 3 > 2\\ &Q_3 = 3!+1 = 7\quad\ \text{has prime factor}\ \ \ \, 7 > 3\\ &Q_7 = 7!+1 = 71^2\ \text{has prime factor}\ \ 71 > 7\\ &\quad\ \ \ \vdots\qquad \qquad \qquad\qquad\ \ \vdots \end{align}\qquad $$ This proof is a minor variation on Euclid's classical proof (which also was not by contradiction, despite many inaccurate historical claims to the contrary).

HINT: If there were only finitely many primes, there would be a largest prime; call it $q$. Now consider what you know about $Q_q$.

Let $P_n$ be a prime dividing $Q_n$. You've proved that $P_n>n$. The following is an infinite sequence of distinct primes: $$\left\{2,P_2,P_{P_2},P_{P_{P_2}},P_{P_{P_{P_2}}},P_{P_{P_{P_{P_{2}}}}},\ldots\right\}.$$ Therefore, there are infinitely many primes.