n! refers to a factorial, a product of n numbers, each one less than the preceding value. You can write a factorial n! by starting with the number n, multiplying it by one less than the previous number, and repeat until you reach 1, at which time you can
stop. So 5! can be written as follows: 5! = 5 * 4 * 3 * 2 * 1 = 120. Similarly, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. Note all the repetition here... it turns out 6! = 6 * 5! That's important because you can re-write (n+1)! as (n+1) * n! So when you divide (n+1)!
by n! as you have in the first term, the n! cancels out in both numerator and denominator leaving only (n+1). So as Robert points out, subtracting n from the first term would leave 1, not 0. So it appears the second term should be (n+1) to be valid.