Verify my proof that $N$ in $ N = (P_1 \cdot P_2...P_n)+1 $ must be odd. I'm a noob. Is the following valid?
$$ N = (P_1 \cdot P_2...P_n)+1 $$ 
where $Pi$ is the $i^{th}$ consecutive prime number and $n$ is any natural number. If $N$ is even, $\dfrac{N}{2} \in \Bbb{N}$, which can be put into the first equation like so:
$$ \frac{N}{2} = \frac{(P_1 \cdot P_2...P_n)}{2}+\frac{1}{2} $$
where it is expected that $ \dfrac{(P_1 \cdot P_2...P_n)}{2}+\dfrac{1}{2} \in \Bbb{N} $ also. $P_1 = 2$, being the first prime number and thus
$$ \frac{(P_1 \cdot P_2...P_n)}{2} = \frac{2(P_2 \cdot P_3...P_n)}{2} = P_2  \cdot P_3...P_n $$
where $P_2 \cdot P_3...P_n \in \Bbb{N}$. Because $\dfrac{1}{2} \notin \Bbb{N}$, $\dfrac{(P_1 \cdot P_2...P_n)}{2}+\dfrac{1}{2} \notin \Bbb{N}$ because any natural number plus a non-natural number is not a natural number.
Because the RHS of the first equation does not belong to $\Bbb{N}$, $\dfrac{N}{2} \notin \mathbb N$ also. Thus $N$ is odd.
 A: That is really complicated, but yes, you are right.
An easier way to see it would be the following:
As $P_1 = 2$, we have that 
$$P_1\cdot P_2 \cdot \ldots \cdot P_n = 2\cdot P_2 \cdot \ldots \cdot P_n$$
is even.
Now $1$ is odd, and you might know that even plus odd always gives odd, case closed. :)
A: If $P_1 = 2$, obviously $N = 2k + 1$ is odd, so your argument after "If $N$ is even.." is invalid.
A: This reminds me of Conway's PRIMEGAME, in which a list of fractions is used to obtain powers of 2 with prime indices. It makes the sieve of Eratosthenes look very sophisticated by comparison. But your question assumes we already know the prime numbers.
You state that $P_1 = 2$. That's obviously even. So is $P_1 P_2$. And $P_1 P_2 P_3$. And any $P_1 P_2 \ldots P_n$ for $n > 2$. These numbers are all divisible by 2 but not by 4: 2, 6, 30, 210, 2310, 30030, 510510, 9699690, 223092870, 6469693230, etc. See Sloane's A002110.
So halving $N$ is an entertaining but unnecessary sidetrack. Since $P_1 P_2 \ldots P_n$ is even, it immediately follows that $N = P_1 P_2 \ldots P_n + 1$ is odd.
This is not to say that there don't exist situations where halving a number is useful. There are such situations. It's just that this one is not such a situation.
