An upper bound for the primorial can be found based on the first chebyshev function.
From $\vartheta(x) < 1.00028x$, it is clear that:
$$\prod\limits_{p \le x}p \le e^{1.00028x} < (2.72)^x$$
I am looking at the following and am having trouble figuring out the upper bound.
$$\prod\limits_{p \le x}p^{\frac{1}{p}}$$
Is there a function similiar to the first chebyshev function that can be used? The second chebyshev function is not helpful since:
$$\psi(x) = \sum\limits_{n=1}^{\infty}{\vartheta(x^{\frac{1}{n}}})$$
I am wondering if there is an upper bound better than a very simple approach such as this:
$$\prod\limits_{p \le x}p^{\frac{1}{p}} < \prod\limits_{p \le x}p^{\frac{1}{2}} < \sqrt{(2.72)^x} < (1.65)^x$$
Like pretty much all products, the way to approach this product is to take logarithms and first estimate $$ \log \prod_{p\le x} p^{1/p} = \sum_{p\le x} \frac{\log p}p. $$ It turns out that this sum is precisely one treated by Mertens: in 1874 he proved that $$ \biggl| \sum_{p\le x} \frac{\log p}p - \log x \biggr| \le 2, $$ which already shows that $\displaystyle e^{-2}x \le \prod_{p\le x} p^{1/p} \le e^2x$.
It turns out that the prime number theorem is equivalent to the assertion $$ \lim_{x\to\infty} \biggl( \sum_{p\le x} \frac{\log p}p - (\log x + A) \biggr) = 0, $$ where $A$ is a constant related to Euler's constant $\gamma$, namely $$ A = -\gamma - \sum_p \frac{\log p}{p(p-1)} \approx -1.33258. $$ We conclude that $\displaystyle\prod_{p\le x} p^{1/p} \sim e^Ax$ where $e^A \approx 0.263795$.