Why product of primes $+1$ is not divisible by any of the product primes? Given $n$ prime numbers, $p_1, p_2, p_3,\ldots,p_n$, then $p_1p_2p_3\cdots p_n+1$ is not divisible by any of the primes $p_i, i=1,2,3,\ldots,n.$ I dont understand why. Can somebody give me a hint or an Explanation ? Thanks.

First approach:

Let $P= 2\times3\times5\times7\times11\times13.$ Then:
The next number after $P$ that is divisible by $2$ is $P+2.$
The next number after $P$ that is divisible by $3$ is $P+3.$
The next number after $P$ that is divisible by $5$ is $P+5.$
The next number after $P$ that is divisible by $7$ is $P+7.$
The next number after $P$ that is divisible by $11$ is $P+11.$
The next number after $P$ that is divisible by $13$ is $P+13.$
So $P+1$ is not divisible by any of those.

Second approach:

If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $2$ then the quotient is $\underbrace{3\times5\times7\times11\times 13}_{\large\text{excluding 2}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $3$ then the quotient is $\underbrace{2\times5\times7\times11\times 13}_{\large\text{excluding 3}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $5$ then the quotient is $\underbrace{2\times3\times7\times11\times 13}_{\large\text{excluding 5}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $7$ then the quotient is $\underbrace{2\times3\times5\times11\times 13}_{\large\text{excluding 7}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $11$ then the quotient is $\underbrace{2\times3\times5\times7\times 13}_{\large\text{excluding 11}}$ and the remainder is $1.$
If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $13$ then the quotient is $2\underbrace{\times3\times5\times7\times11}_{\large\text{excluding 13}}$ and the remainder is $1.$
(Appendix: $(2\times3\times5\times7\times11\times13) + 1 = 59\times509.$)

