Is the product of all primes a natural number? In other words, is this true:
$$ \prod\limits_{\text{primes}} p_i \in \mathbb{N} $$
And if so, what about just some of them:
$$ \prod\limits_{\overset{p_i \text{ are primes}}{p_i = 9(mod 10)}} p_i \in \mathbb{N} $$
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Proof by contradiction, suppose that the product of is indeed natural number,
$$\prod\limits_{\text{primes}} p_i =k\in \mathbb{N}$$
thus from the fundamental theorem of arithmetic we have an index $l\in \mathbb{N}$ for which the set $\{p_1,p_1,\dots,p_l \}$ contains all prime numbers, but there are infinite many of them. Therefore $k$ is not a natural number.
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If the product were a natural number $N$, there would be only finitely many primes, and ask yourself which primes $N\pm1$ could possibly be divisible by, when $N$ is divisible by every prime.
(The simpler way than that to prove that there are infinitely many primes is this: consider any finite set of primes (not assumed to contain all primes); multiply them and add or subtract $1$. Show that the prime factors of the number you get cannot be in the finite set you started with. So there are more primes than can be found in any finite set.)
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In a sense there are "too few" primes in the product you propose for it to be a well-defined notion, but if you consider the product of all primes smaller than a fixed infinite hypernatural $H$, this does give you a well-defined notion: $$\prod_{p\leq H,\; p\;\text{prime}}p\quad\quad\quad\quad\quad (*)$$ To express this is a slightly more precise (but also more technical) form, the set $\mathbb{N}$ inside the hypernaturals is not an internal set and therefore you cannot form the precise product you envisioned.
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One can make sense of some divergent series by using zeta regularization techniques. In the case of infinite product of prime numbers, this paper http://link.springer.com/article/10.1007%2Fs00220-007-0350-z provides the surprising answer $4 \pi^2$.
The answer to both your questions is NO.
Dirichlet's theorem states that there are infinitely many primes of the form $a\pmod{d}$, where $\gcd(a,d) = 1$.
Hence, the product of infinitely many numbers greater than $2$ is not a natural number, i.e., if $\gcd(a,d) = 1$, then $$f(n) = \prod_{\overset{p \equiv a \pmod d}{p = \text{prime}\leq n}} p$$then $\lim_{n \to \infty} f(n) = \infty$.