We give two proofs. The first is probably the intended one. The second is perhaps more interesting, and certainly more informative. For one thing it shows that there are infinitely many primes in a way quite different from the standard proof that goes back to Euclid's Elements.
First proof: Note that $p_{k+1}\le (p_1p_2\cdots p_k)+1$, since some prime $p$ divides $(p_1p_2\cdots p_k)+1$, and $p$ cannot be any of $p_1,p_2,\dots,p_k$. Now use the induction hypothesis. We have $p_1p_2\cdots p_k\le 2^{e_k}$ where $e_k \le 2^0 +2^1+\cdots +2^{k-1}.$ The series on the right is a geometric series with sum $2^k-1$. So $p_1p_2\cdots p_k <2^{2^k}$, and therefore $(p_1p_2\cdots p_k)+1 \le 2^{2^k}$.
Second proof: Let $n \ge 1$. We will show that $2^{2^n}-1$ has at least $n$ distinct prime factors. None of these prime factors is equal to $2$. So there are at least $n+1$ primes that are $\le 2^{2^n}$.
We prove that $2^{2^n}-1$ has at least $n$ prime factors by induction on $n$. Suppose that we know that for a particular $k\ge 1$, there are at least $k$ primes that divide $2^{2^k}-1$. We want to show that there are at least $k+1$ primes that divide $2^{2^{k+1}}-1$.
We use the familiar factorization (difference of two squares) $2^{2^{k+1}}-1=(2^{2^k} -1)(2^{2^k} +1).$ By the induction hypothesis, $2^{2^k} -1$ has at least $k$ prime factors. Let $p$ be a prime factor of $2^{2^k}+1$. Then $p$ cannot divide $2^{2^k} -1$, since if it did, it would divide the difference $(2^{2^k} +1)-(2^{2^k} -1)$, which is $2$. But since $2^{2^k}+1$ is odd, $p$ must be odd. So in addition to the $k$ prime factors of $2^{2^k} -1$, the number $2^{2^{k+1}} -1$ has at least one additional prime factor $p$, for a total of at least $k+1$.
Comment: The formal induction hides the simplicity of the idea. It all comes down to the fact that, for example, $2^{2^5}-1=(2^{2^1}-1)(2^{2^1}+1)(2^{2^2}+1)(2^{2^3}+1)(2^{2^4}+1).$