33 = 27 < 3!2 = 62 = 36.

Bzzt! Wrong answer!

Not true. nn ≤ n!2 for all n a natural number.


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

This sequence contains n terms and each term is of the form ((n-i)*(i+1))
(n-i)*(i+1) = ni - i2 + n - i
This is obviously greater than or equal to n for all i such that 1 ≤ i ≤ n
Hence, the sequence is the product of n terms, all greater than n. Hence it is greater than or equal to nn.


Jurph: Fixed. Now it says ≤ all the way through, rather than <. Stupid typos.
...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 - i2 + 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 nn = 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, nn 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.

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