...Except that i isn't restricted to the domain (1...i...n).

Where above, you said "the sequence is the product of n terms, all greater than n," you should have said, "the sequence is the product of n terms, all greater than *or equal to* n." The number of terms equal to n (2 of them) comes into play for extreme values of i.

You have to start your sequence with i=0 and end with i = (n-1) for your end-terms to be equal to (n*1) or (1*n), as follows: For i = (n-1),

(n - i) * (i + 1) = ni - i

^{2} + n - i

= 1 * n = n.

Likewise for i=0, the first term = n.

Since your first and last terms are equal to n, and you have n terms, you still need to prove that just one of those intervening terms is strictly greater than n to have solid proof.

If you only have 2 terms (and 2 is a natural number), the argument fails either way, and n^{n} = n!^{2}. That is, you can't prove one is greater or less for all cases, because of the equality case.

For n, any

natural number, n

^{n} and n!

^{2} have no provable relationship. Although your proof

*does* refute the theory for all cases n > 2, you need this little caveat tacked on for n = 2. The trivial case n = 1 also disproves the theory above, but that is left as

an exercise for the reader.

Okay, the author above fixed a typo from "less than" to "less than or equal", so this mostly goes in /dev/null. Just a proof of the trivial case.