Is there an elementary proof that for primes $p_i$,
$$\lim_{n \to \infty} \prod_{i=1}^n \left(1-\frac{1}{p_i}\right)=0\;?$$
This follows from Mertens third theorem which states that
$$ \prod_{i=1}^n \left(1-\frac{1}{p_i}\right) \sim \frac{e^{-\gamma}}{\log{n}} $$
but I was hoping for a self contained proof.
As indicated in a comment, this is as elementary to show as $\sum 1/p = \infty$.
For all real numbers $x$ we have $e^x \geq 1+x$ (note $y = 1+x$ is the tangent line to $y = e^x$ at $(0,1)$), so $\boxed{e^{-1/p} \geq 1 - 1/p}$. Therefore $$ 0 < \prod_{p \leq t} \left(1 - \frac{1}{p}\right) \leq e^{-\sum_{p \leq t} 1/p}. $$ So as soon as you know that $\sum_{p \leq t} 1/p \to \infty$ as $t \to \infty$, you know the right side of the displayed inequality above tends o $0$ as $t \to \infty$, so $\prod_{p \leq t} (1-1/p) \to 0$ as $t \to \infty$.