I want to show that the limit of the geometric mean of primes less than or equal to $x$ is $e$ as $x \to \infty$. Is this correct?
Using the product law of logarithms we have $\ln \prod\limits_{p \leqslant x} p = \sum\limits_{p \leqslant x} {\ln (p)}$ but $\sum\limits_{p \leqslant x} {\ln (p)} = \vartheta (x) \sim x$ hence $\prod\limits_{p \leqslant x} p \sim {e^x}$ and the geometric mean is ${\left( {\prod\limits_{p \leqslant x} p } \right)^{1/x}} \sim e.$
Edit: I realized that this product does not represent the geometric mean. What does it represent?