$n=2012^{2011^{2010}}$

I'm thinking that you first find

$n_1=2012^{2011}\pmod{10^{10}}$

and then

$n_2 = n_1^{2010} \pmod{10^{10}}$

and if say $10^{10}$ was prime, or relatively prime to $2012$ then Fermat's Little Theorem or Euler's Theorem can be brought into play but that doesn't occur here.

Is there some trick I'm not aware of? (almost certainly)