How to solve the recurrence relation $f(1,n)=n+2$? Given the following information:
$f(0,n)= n+1$ $ \ \ $ $\forall n$
$f(m,0)=f(m-1,1)$ when $m>0$
$f(m,n)=f(m-1, f(m,n-1))$ when $m>0$ and $n>0$ 
I have worked out that $f(1,n)=f(0,f(1,n-1))= f(1,n-1) + 1$
But I am unsure of how to get $f(1,n) =n+2$ from this stage ? 

\begin{eqnarray*}
f(1,0)=f(0,1)=2 \\
f(1,1)=f(0,f(1,0))=f(0,2)=3 \\
f(1,2)=f(0,f(1,1))=f(0,3)=4 \\
\ddots
\end{eqnarray*}
