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Suppose n is a natural number,
Prove that is always an integer.
Suppose there are n consecutive natural numbers starting with m.
product = m(m+1)(m+2).....(m+n-1) = (m+n-1)!/(m-1)! = n!* m+n-1Cn
We know m+n-1Cn is an integer. Therefore any n consecutive integers are divisible by n!
Now we take (n!)!. This means product of all numbers from 1 to n!
We divide this product into rows containing n consecutive integers.
1 2 3 4 ............ n
n+1 n+2 n+3 n+4 .............. 2n
2n+1 2n+2 2n+3 2n+4 ........... 3n
.
n!-n+1 n!-n+2 n!-n+3 ............. n!
There are (n-1)! rows each containing n numbers [ n(n-1)! = n! ]
Therefore, product of numbers in each row is divisible by n!
Therefore the total product is divisible by (n!)(n-1)!
So, (n!)!/(n!)(n-1)! is an integer.
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