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.
 A: Towards a contradiction, if some prime $p_i$ ($1\leq i\leq n$) divides $p_1p_2p_3\cdots p_n+1$, then because  $p_i$ also divides $p_1p_2p_3\cdots p_n$, it follows that $p_i\mid 1$ (the difference), a contradiction. 
A: 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: $(2\times3\times5\times7\times11\times13) + 1 = 59\times509.$)

A: Suppose it divide bt $p_{i}$. Then $n \equiv 0 \mod p_{i}$. 
But $n = p_{1} \dots p_{n} + 1 \equiv 1 \mod p_{i}$.
A: Simple example.  Suppose I consider $2 \cdot 3 \cdot 5 \cdot 7 = 210$
Now, $2$ divides $210$, and so do $3$, $5$, and $7$ ... of course!  
But what happens if you divide $210+1=211$ by $2$?  You get a remainder of $1$ ... exactly because you got a remainder of $0$ when dividing $210$.  And the exact same thing happens for $3$, $5$, and $7$ 
A: If some integer $m$ is divisible by e.g. $13$ then $m+1$ is not.
Now note that $m=p_1p_2\cdots p_n$ is divisible by every $p_i$ with $i\in\{1,\dots,n\}$ and draw conclusions.
A: Because if $k>1$ divides $n$ then $k$ can not divide $n+1$.  
If $n = m*k$ then then $n + 1 = k(m + \frac 1m)$ and $m + \frac 1m$ is not an integer.  
Or to put it another way.  If $n = m*k$ then the next multiple of $k$ is $n+k = m*k + k$ which is larger than $n+1 = mk + 1$.
So because $p_i$ divides $p_1p_2.....p_n$ and $p_i > 1$ then $p_i$ can not divide $p_1p_2.....p_n + 1$ because $p_1p_2.....p_n + 1 = p_i(p_1.....p_{i-1}p_{i+1}... p_n + \frac 1{p_i})$ and $p_1.....p_{i-1}p_{i+1}... p_n + \frac 1{p_i}$ is not an integer.
!!!!VITALLY!!!!  IMOPORTANT POST-SCRIPT:  This does NOT mean $p_1p_2.....p_n + 1$ is prime!  It just means that none of the $p_i$ for $i \le n$ divide it.  It is possible that there is a $p_m; m > n$ that $p_m$ does divide $p_1p_2.....p_n + 1$.  This is not a contradiction because $p_m$ does not divide $p_1p_2.....p_n$.
A: Simpler argument is that dividing by any $p_k$ we get the remainder $1$.
A: Let $$P = p_1p_2...p_n+1$$ and let $p$ be a prime such that $p\mid P$.
Then $p$ can not be any of $p_1,p_2,p_3,\cdots ,p_n$ otherwise $p$ would divide the difference $P-p_1p_2...p_n=1$ which is not possible.
