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 $11\cdot 13 +1=12^2$.
Let $P\_n = \underbrace{\color{red}{p\_1}p\_2 \cdots p\_n}\_{Q\_n} +1$. As $\color{red}{p\_1}=2$ and $2 \nmid p\_i$ for $i≥2$ therefore $Q\_n \equiv 2$(mod $4$) $\implies P\_n \equiv 3$(mod $4$). But for $n \in \mathbb{Z}, n^2 \not \equiv 3$(mod $4$). Therefore $P\_n$ is not a perfect square. $\Box$