The digit in the one's place of the number 
corresponds to:
I can see that this problem isn't difficult at all, yet I would like to know the reasoning behind its answer (which is 9). I first thought I could try and decompose the exponent to 
, yet I dont really see how this would help me. Can someone please explain how to get the answer for me?
Lets look at the one's place of various powers of 23: $$ \begin{align} 23^1&\equiv3\pmod{10}\\ 23^2&\equiv9\pmod{10}\\ 23^3&\equiv7\pmod{10}\\ 23^4&\equiv1\pmod{10}\\ 23^5&\equiv3\pmod{10}\\ \end{align} $$ We can see that the one's digit has a period of 4. Now we want to cycle the list. Since the one's digit repeats every 4, we can do $$ 2014\equiv2\pmod{4}. $$ The answer is the second one in our list, 9.