#### SB257426

##### Very Important User

By way of contradiction assume that 2k is the largest even integer.

Now consider (2k)!

(2k)! = (2k)(2k-1)(2k-2).......(k)(k-1)(k-2)......(2)(1)

= 2[k(2k-1)(2k-2).......(k)(k-1)(k-2)......(2)(1)]

= 2p, which is also an even integer. This contradicts the fact that 2k was the largest even integer. Therefore it is sensible to claim that there is no even integer since 2p>2k