\(\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1}{x^2+5x+5}\)
=\(\dfrac{\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1}{x^2+5x+5}\)
=\(\dfrac{\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)+1}{x^2+5x+5}\)
=\(\dfrac{\left(x^2+5x+5\right)^2-1+1}{x^2+5x+5}\)=1