Here is an alternative method which doesn't require factoring a fourth degree polynomial. We start with n(n+1)(n+2)(n+3), and since the order of the factors don't matter we can say that

n(n+1)(n+2)(n+3) = n(n+3)(n+1)(n+2).

Focusing on the right hand side we multiply the first and second factors as well as the third and fourth to get

n(n+1)(n+2)(n+3) = (n^2+3n)(n^2+3n+2).

Which can be rewritten as

n(n+1)(n+2)(n+3) = [(n^2+3n+1)-1][(n^2+3n+1)+1].

This is a difference of squares, hence

n(n+1)(n+2)(n+3) = (n^2+3n+1)^2-1.