The square of n+1-th prime is less than the product of the first n primes. I wanted to prove the following question in an elementary way not using Bertrand postulate or analytic estimates like $x/\log x$. The question is 
$$
p_{n+1}^2<p_1p_2\cdots p_n,\qquad(n\geq4)
$$
I made the following argument. Does anyone have some opinion or simpler ideas to complete.
We consider two cases:
Case 1: $N=p_1p_2\cdots p_n-1$ is composite: then there will be a prime factor  $q\leq\sqrt{N}$, and of course $q$ is not any of the $p_i$'s. therefore $p_{n+1}\leq q\leq\sqrt{N}$. So $p_{n+1}^2<N+1$.
Case 2: $N=p_1p_2\cdots p_n-1$ is prime. Then???
 A: The inequality 
$$p_{n+1}^2<p_1p_2\cdots p_n$$
for $n>3$  is called Bonse inequality, and Bonse gave a completely elementary proof in
 Bonse, H. (1907). "Über eine bekannte Eigenschaft der Zahl 30 und ihre Verallgemeinerung". Archiv der Mathematik und Physik 3 (12): 292–295. It is also proved in the book of Uspensky and Heaslet, Elementary Number Theory,
on pages 87-89. 
An online solution can be found at art of problem solving. A further online proof is in this book, Theorem 13.
Basically the proof uses 
$$
n-m+1<p_m
$$
for $m=[n/2]$, and $p_{n+1}<p_1\cdots p_m$. Then
\begin{align*}
p_{n+1}^2 & < (p_1\cdots p_m)(p_1\cdots p_m) \\
 & < (p_1\cdots p_m)(p_{m+1}p_{m+2}-p_{2m})\\
 & < p_1\cdots p_n.
\end{align*}
A: Hint: try to do an induction on the primes numbers, where
$$p_1=2,p_2=3,p_3=5...$$
And use the theorem saying that 
$$p_{n+1}\leq 2p_n.$$
A: Consider $a = p_1 p_2 p_3 \dots p_n p_{n+1}$
All the prime divisors of $a$ must be less than $< a^{\frac{1}{2}}$
Therefore $p_{n+1} < a^{\frac{1}{2}} = (p_1 p_2 p_3 \dots p_{n} p_{n+1})^{\frac{1}{2}}$
Conclusion : $(p_{n+1})^2 < p_1 p_2 p_3 \dots p_n p_{n+1}$
