We already have $p_n \gt 2(n-2)$ for $n \gt 0$ by elementary (non-analytic) methods. Similar methods can establish the inequality for sufficiently large n.
Suppose the inequality $1 \lt A \prod_{i=3}^n \frac{p_i - 2}{p_i}$ holds for positive integers $n=j$ and $j+1$. I will show the inequality holds for $j+2$, leaving the inductive conclusion to the reader. Note that the term for $j+2$ is at least the term for $j$ times $B=\frac{(A+1)(p_{j+1}-2)(p_{j+2}-2)}{Ap_{j+1}p_{j+2}}$. This is because $p_{j+2} \geq 6 + p_j$, so the value of $A$ for the two terms increases by at least 1. But $\frac{A+1}{A}\geq 1 + 6/p_j$ while $(1 - 2/p_{j+1})(1- 2/p_{j+2})\gt 1 - 4/p_j)$ giving $B \gt 1$ and thus the inequality holds for $j+2$.
One can generalize this replacing $A$ by smaller values like $cp_n$ for some positive real $c$ and replacing $p_i-2$ by $p_i-k$ for a larger fixed value of $k$. The rewritten inequality will fail for small $n$ but eventually hold, as analytic methods will have the right hand side grow like a constant times $p_n^2/\log p_n$.
Gerhard "Should Anticipate The Next Questions" Paseman, 2016.03.30.