Let p be prime. Assume (1): $\hspace{10mm} (\prod_{p\leq n} p)^{1/n} \sim e.$
Then $(e^{\ln \prod p})^{\frac{1}{n}} = e^{(\sum \ln p)/n} \sim e \implies \lim_{n=1}^\infty \frac{e^{(\sum \ln p)/n}}{e} = 1. $
And so
$ \lim_{n=1}^\infty~ e^{(\frac{\sum \ln p}{n}-1)} = 1$ or
$\lim \frac{\sum \ln p}{n } - 1 = 0 \implies \lim (\frac{\vartheta(n)}{n}- 1) = 0$
But this implies that $\lim_{n = 1}^\infty~ (\vartheta(n) - n) = 0 $
which is false.
I have good reason to think (1) is true so perhaps someone can point to the error, which I will chalk up to hurricane-fatigue.
Thanks!