Factoring $(x+1)(x+2)(x+3)(x+4)+1$ I have to factor this polynomial:
$$(x+1)(x+2)(x+3)(x+4)+1$$
WolframAlpha gives 
$$(x^2+5x+5)^2$$
How can I prove it without expanding the result?
Thanks!

The key for this problem is observing that $(x+1)(x+2)(x+3)(x+4)+1$ is symmetric with respect to $x=-2.5$. Write it as
$$(x+1)(x+2)(x+3)(x+4)+1= \\
 \left( (x+2.5)-1.5\right)\left( (x+2.5)-0.5\right)\left((x+2.5)+0.5 \right)\left((x+2.5)+1.5 \right)+1$$
For simplicity, let us denote temporarily $x+2.5$ by $y$. Then we need to factor
$$\left( y-1.5\right)\left(y+1.5 \right)\left(y-0.5\right)\left(y+0.5 \right)+1=\left( y^2-1.5^2\right)\left(y^2-0.5^2\right)+1$$
Now, this is a quadratic in $y^2$, which can be factored by completing the square. Then replace $y=x+2.5$. 
