Yes, there is a pattern. The fraction converges to $e$.
This is essentially a consequence of the prime number theorem (in a sufficiently precise form). [Sorry for the initial totally messed up version, where I also overlooked something; and while the answer below is not hard either it is slightly more subtle than I thought at first, so first no details, now too many...so on average alright.]

Let $p_n$ denote the $n$th prime.

The question asks for the quantity

$$
\frac{p_{n+1}}{ (\prod_{i=1}^n p_i)^{1/n} }
$$

Now, $\log(\prod_{i=1}^n p_i)$ is Chebychev's Theta of $p_n$ and as such well-studied.
In partiular one knows (see for instance Dusart's thesis page 15) that it is $n (\log n + \log \log n - 1 + o(1))$.
So the denominator in the expression above is

$$
\exp (\log n + \log \log n - 1 + o(1)) = n (\log n) e^{-1} exp(o(1))
$$

while the numerator is $p_n \sim n (\log n)$.

Therefore the fraction converges to $e$.