Yes, your approach is right. You can also solve it "recursively" with Euler's theorem, which says that for any $a,x,y,k,$
$x \equiv y \bmod \phi(k) \ \Rightarrow \ a^x \equiv a^y \bmod k.$
In this case you want to find $A = 2013^{2012^{\ldots}} \bmod 10$ and $B = 2012^{2011^{\ldots}} \bmod 10$. For $A$ we apply the above with $k = 10$ and $\phi(k) = 4$ to get:
$2012^{2011^{\ldots}} \equiv 0 \bmod 4 \ \Rightarrow \ 2013^{2012^{\ldots}} \equiv 2013^0 \equiv 1 \bmod 10.$
Applying the above twice, starting with $k = 10$ (so that $\phi(k) = 4$ and $\phi(\phi(k)) = 2$), for $B$ we get
$2010^{2009^{\ldots}} \equiv 0 \bmod 2 \ \Rightarrow \ 2011^{2010^{\ldots}} \equiv 2011^0 \equiv 1 \bmod 4 \ \Rightarrow \ 2012^{2011^{\ldots}} \equiv 2012^1 \equiv 2 \bmod 10.$
In other words:
$\displaystyle a^{\displaystyle b^{\displaystyle c^{\displaystyle d^{\ldots}}}} \equiv (a \bmod k)^{\displaystyle (b \bmod \phi(k))^{\displaystyle (c \bmod \phi^2(k))^{\displaystyle (d \bmod \phi^3(k))^{\ldots}}}} \bmod k,$
where e.g. $\phi^3(k) = \phi(\phi(\phi(k)))$. Applying this to $A$ we get
$2013^{\displaystyle 2012^{\ldots}} \equiv (2013 \bmod 10)^{\displaystyle (2012 \bmod 4)^{\ldots}} \equiv 4^{\displaystyle 0^{\ldots}} \equiv 1 \bmod 10,$
and for $B$ we get
$2012^{\displaystyle 2011^{\displaystyle 2010^{\ldots}}} \equiv (2012 \bmod 10)^{\displaystyle (2011 \bmod 4)^{\displaystyle (2010 \bmod 2)^{\ldots}}} \equiv 2^{\displaystyle 3^{\displaystyle 0^{\ldots}}} \equiv 2 \bmod 10.$