Help, I'm Trapped in a Math Factory!

After the $2015$ th sorting, Alex has $2(1)+1 = 2 + 1$ numbers. (This is basically the inverse of the given function $f(x) = \frac{1}{2}(x-1)$ in the problem.) After the $2014$ th sorting, Alex has $2(2(1)+1)+1 = 2^{2} + 2 + 1$ numbers, and so on. Thus, after the $2016-n$ th sorting, where $n$ is a positive integer less than $2016$ , there are $2^{n} + 2^{n-1} + \cdot + 2 + 1$ numbers left. For $n = 2016$ , i.e. the start, we have $2^{2016} + 2^{2015} + \cdots + 2 + 1 = 2^{2017} - 1,$ where we apply the geometric series formula to calculate the series. $2^{2017}$ has a units digit of $2$ , so $2^{2017} - 1$ has units digit $\boxed{1}.$