Question: Can you please break this proof downto a level where a high school student can understand it? Thankyou.

Example:

Prove that the product of n successive integers is alwaysdivisible by n!

If the product of n successive integers is always divisible byn!, then we would only need to prove it is true for positiveintegers. If one of the integers in the product is zero, it willalways be true. If the integers are negative, n! will divide bytheir absolute value.

Proof by contradiction:

If there is a number of n successive positive integers whoseproduct is not divisible by n!, then we can choose the smallest andcall it N. N must be greater than 2 because the product of any twosuccessive integers is always even. Therefore, there must be aninteger, m, such that (m+1)(m+2)...(m+N) is not divisible by N! Ofthese numbers, m, let M be the smallest. M must be positive sinceN! is divisible by N!. So, we are supposing that (M +1)(M+2) …(M+N)is not divisible by N!

(M+1)(M+2) …(M+N-1)(M+N) =M[(M+1)(M+2)...(M+N-1)]+N[(M+1)(M+2)...(M+N-1)]

Using our choice of M, n! divides into M[(M+1)(M+2)...(M+N-1)].Using our choice of N, (N-1)! divides (M+1)(M+2)...(M+N-1) andtherefore N! divides N[(M+1)(M+2)...(M+N-1)].

In combination, N! Divides the right side of the last equation.This contradiction establishes the result.