Solve the equation n! + (n+1)! = (n+2)!.

(n+2)!= 1*2*3*...(n-1)*(n)*(n+1)*(n+2) = n!*(n+1)*(n+2)

(n+1)!= 1*2*3*...(n-1)*(n)*(n+1) = n!*(n+1)

n!= 1*2*3*..*(n-1)*n

n! + (n+1)! = (n+2)!

We'll re-write the equation:

n! + n!*(n+1) = n!*(n+1)*(n+2)

We'll factorize, to the left side:

n!*(1+n+1) = n!*(n+1)*(n+2)

n!*(n+2) = n!*(n+1)*(n+2)

We can divide by the same factors, from both sides.

1 = n+1

We'll add the value -1, both sides:

1-1 = n+1-1

n=0, is the solution of the equation.