smc

Simply take some time to draw a table of values of $f(i,j)$ for the first few values of $i$: $$\begin{array}{cccccc}i\text{ }j& 0& 1& 2& 3& 4\\ 0& 1& 2& 3& 4& 0\\ 1& 2& 3& 4& 0& 1\\ 2& 3& 0& 2& 4& 1\\ 3& 0& 3& 4& 1& 0\\ 4& 3& 1& 3& 1& 3\\ 5& 1& 1& 1& 1& 1\end{array}$$ Now we claim that for $i\ge 5$, $f(i,j)=1$ for all values $0\le j\le 4$. We will prove this by induction on $i$ and $j$. The base cases for $i=5$, have already been proven. For our inductive step, we must show that for all valid values of $j$, $f(i,j)=1$ if for all valid values of $j$, $f(i-1,j)=1$. We prove this itself by induction on $j$. For the base case, $j=0$, $f(i,0)=f(i-1,1)=1$. For the inductive step, we need $f(i,j)=1$ if $f(i,j-1)=1$. Then, $f(i,j)=f(i-1,f(i,j-1)).$ $f(i,j-1)=1$ by our inductive hypothesis from our inner induction and $f(i-1,1)=1$ from our outer inductive hypothesis. Thus, $f(i,j)=1$, completing the proof. It is now clear that for $i\ge 5$, $f(i,j)=1$ for all values $0\le j\le 4$. Thus, $f(2015,2)=\boxed{1}$.