Powers of 2

If you wanted to look at the pattern, the last digit always repeats 2,4,8,6,2,4,8,6... And will never be 0

$2$ is an even number. That is, if $2$ is raised to any integer exponent, then it would yield an even numbered value. If $2^x$ yields a number ending in zero for some integer $x$ , then one half of this number, i.e, $2^{x-1}$ , is some integer ending in $5$ or $0$ . But as we established earlier, if $2$ is raised to any integer exponent, then it would yield an even numbered value. So $2^x$ can never end in $5$ . If $2^x$ was some number ending in zero, then so would be $2^{x+1}$ , and every other exponent of $2$ that is greater than $x$ . Therefore, there is no integer $x$ for which $2^x$ yields a number ending in zero.

2^x contains all 2's. In order to end in 0 would require the presence of a 5. Ed Gray

For a number to end in a 0 it will need at least a prime factor of 2 and 5 and as 2^{x} will never have a prime factor of 5 it will never end in 0