Primes of the form $p_1p_2\dotsm p_k+1$ Let $p_1,p_2,...,p_k$ be the first $k$ primes and $k>1$. Does there exist a prime $p_{k+1}$ such that $p_{k+1} = p_1p_2...p_k + 1$?
 A: Bertrand's postulate states that there is always a prime between $n$ and $2n$, whenever $n > 3$. Since $p_1 = 2$, you are asking for instances when $p_{k+1} = 2 p_2 \dots p_k + 1$; but we know $p_{k+1}$ lies between $p_k$ and $2 p_k$, so in particular it is less than $2 p_k$ and hence less than $2 p_2 \dots p_k + 1$.
A: There only exists one such prime, and that's 3, because by Bertrand's postulate we see that... oh, too slow, Patrick already mentioned Bertrand.
I'm going to try another way. The prime number theorem tells us that $$\pi(n) \sim \frac{n}{\log n}.$$ It's not very precise for small numbers, but it will do for our purposes here.
So, if $$\pi\left(\prod_{i = 1}^{k - 1} p_i\right) \approx \frac{\prod_{i = 1}^{k - 1} p_i}{\log \prod_{i = 1}^{k - 1} p_i}$$ and $$\pi(p_k) \approx \frac{p_k}{\log p_k},$$ then $$\pi\left(\prod_{i = 1}^{k} p_i\right) \approx \frac{\prod_{i = 1}^{k} p_i}{(\log p_k) + \log \prod_{i = 1}^{k - 1} p_i}.$$
Remember the basic properties of logarithms: we have a multiplication for the numerator and an addition for the denominator.
And since $\log p_k > 1$ for $k > 1$, the denominator does increase but not as much as the numerator. This storngly suggests that there are at least a couple of primes between $$\prod_{i = 1}^{k - 1} p_i$$ and $$\prod_{i = 1}^{k} p_i,$$ so even if $$1 + \prod_{i = 1}^{k} p_i$$ is indeed prime, it can't be the next prime.
Try it out with, say, the first four primes and a scientific calculator.
