Let $p_i$ ($i \in \mathbb{N}^+$) be the $i^\text{th}$ prime. Is the product of the first $k$ primes always strictly greater than $p_{k+1}^2$ when $k > 3\text{?}$ (For instance, for $k=4,$ $2\cdot 3\cdot 5 \cdot 7 = 210 > 11^2\text{.)}$ If so, how would one prove it?
I'm not sure how I would use induction, because there are arbitrarily large prime gaps, but not much is known about where they are in the sequence of primes, other than that a gap of size $n$ can always be found starting at $n!.$
On
Here's what I think is a simpler approach than the answer by Mike. We proceed by induction. Suppose for some $k > 4$
$\displaystyle\prod_{i=1}^{k-1} p_i > p_k^2.\tag*{}$
The base case, $k=5$, can be verified by direct computation. Now we note that $p_n > 4$ for $n > 4$. We multiply the above inequality by $p_k$ on the left and $4$ on the right, which preserves the inequality:
$\displaystyle\prod_{i=1}^k p_i > 4p_k^2.\tag*{}$
By Bertrand's postulate, $p_{n+1} < 2p_n$ or $2 > p_{n+1}/p_n$. Thus
$\displaystyle\prod_{i=1}^k p_i > \left(\frac{p_{k+1}}{p_k}\right)^2p_k^2 = p_{k+1}^2\tag*{}$
which completes the induction step.
For $k \ge 7$ , yes.
There are only $6$ primes less than $16$. So on the one hand, the product $\prod_{i=1}^k p_i$ of the first $k$ primes for $k \ge 7$ is at least $$(2 \times 3 \times 5 \times 7 \times 11 \times 13) \times 16^{k-6}$$ $$> 4^6 \times 16^{k-6}.$$ By Bertrand's postulate on the other hand [which has been established], $p_{k+1} \le 2^{k+1}$ and so $p^2_{k+1} \le 4^{k+1}$.
But the inequality $$\prod_{i=1}^k p_i$$ $$\ge (2 \times 3 \times 5 \times 7 \times 11 \times 13) \times 16^{k-6}$$ $$> \ 4^6 \times 16^{k-6}$$ $$\ge 4^{k+1} \ge p^2_{k+1}$$ holds for all $k \ge 7$.