Question: how to prove $\forall n$ : $p\_1p\_2...p\_n +1$ isn't the perfect square. $p\_i$ is $i$th prime number.

Answer: 
Let $n \ge 1$. Since $2$ is the first prime, our product is even. Thus $p\_1\cdots p\_n+1$ is odd. Suppose it is equal to the odd perfect square $m^2$. 

Note that $m^2-1$ is divisible by $4$, indeed by $8$. But $p\_1\cdots p\_n$ is not divisible by $4$. Thus $p\_1\cdots p\_n+1$ cannot be a perfect square.

**Remark:** The prime $2$ played a crucial role in this argument. Note for example that $3\cdot 5+1=4^2$, and $11\cdot 13 +1=12^2$.
Hints:

$$(1)\;\;\;\;\;\forall\,n\in\Bbb Z\;\;,\;\;\;n^2=0,1\pmod 4$$

$$(2)\;\;\;\;\;\;\;\text{What Thomas Andrews wrote in his comment above}$$