# And one more similar one…

Take any number $n_1>1$. It’s divisible by at least one prime. Since it has no common factors with $n_1+1$, then the new number $n_2=n_1(n_1+1)$ is divisible by at least two primes. Then $n_3=n_2(n_2+1)$ is divisible by at least 3 primes by similar reasoning. In general, $n_k$ is divisible by at least $k$ distinct primes. Since $k$ can grow arbitrarily large, there must be arbitrarily large primes.

This proof has a similar vibe to the first two I gave. In fact, if you unravel the topological proof I gave yesterday, you’ll see even that has its similarities. I promise, we’re going to move on to genuinely different things now.