All even numbers can be represented by 2k, where k is an integer.
The product of a power of 2 (2ⁿ) and an odd number (2m+1), where integer n>0 and m is any integer.
Consider 2ⁿ(2m+1)=2k, 2ⁿ⁻¹(2m+1)=k. When n=1, k=2m+1, which means k is odd. When n>1, k must be even. So for all n>0 we can find any k, that is, any integer. Therefore all even numbers are the product of a power of 2 and an odd number.
There is one exception if zero is accepted as an even number. 2ⁿ(2m+1)=0 has no solutions for m and n, because 2ⁿ≠0 for any n. And 2m+1≠0, because m has to be an integer (m≠-½).