Euler Fermat with double exponent I have to calculate
$$ 3^{{2014}^{2014}} \pmod {98} $$
(without calculus). I want to do this by using Euler/Fermat. 
What I already have is that the $\gcd(3, 98) = 1$ so I know that I can use the Euler Fermat formula. 
Then I know that $\varphi(m = 98) = 42 $
Then I can say that
$$3^{{2014 }^{2014} \pmod {42}} \pmod {98}$$
Now I don't know how to progress. Any ideas/hints? 
 A: Continue for $2014^{2014} \pmod{42} \equiv (2014 \pmod {42})^{2014} \equiv (-2)^{2014} \equiv 2^{2014} $
Chinese Remainder Theorem and Fermat's Little Theorem: $42 = 2 \times 3 \times 7 $
$ 2^{2014} \pmod 2 \equiv 0 \bmod 2 $
$ 2^{2014} \pmod 3 \equiv (-1)^{2014} \equiv 1 $
$ 2^{2014} \pmod 7 \equiv 2^{2014 \bmod \ 6} \equiv 2^4 \equiv 16 \equiv 2 $
CRT: $2^{2014} \pmod{42} \equiv 16 $
We are left with $3^{16} \pmod {98} $
Apply repeated squaring:
$ 3^{16} \equiv (3^4)^4 \equiv (-17)^4 \equiv 17^4 \equiv (17^2)^2 \equiv (-5)^2 \equiv \boxed{25} $
A: Well, I had written almost all of this up, and then @GohP.iHan posted a much better/shorter approach. :)  I'll go ahead and post in case someone else finds it useful, but you should use his way.
Your problem, now, is that you want to compute $2014^{2014}\pmod{42}$.  We can factor 2014, which should make it easier:
$$2014^{2014} = (2\cdot19\cdot53)^{2014}$$
So, we can use Euler/Fermat for $19^{2014}$ and $53^{2014}$:
\begin{align}
19^{2014}&\equiv 19^{2014\bmod \phi(42)}\pmod{42}\\
&\equiv 19^{2014\bmod 12}\\
&\equiv 19^{10}\\
&\equiv \left(19^{-1}\right)^2\equiv 31^2\\
&\equiv 37\pmod {42}
\end{align}
Note how I used that $9^{10}\equiv 19^{12}\cdot \left(19^{-1}\right)^2$; this saves me from having to do repeated-squares to evaluate the power.  I use the same trick for $53^{2014}$ below:
\begin{align}
53^{2014}&\equiv 11^{10}\pmod{42}\\
&\equiv \left(11^{-1}\right)^2\equiv 23^2\\
&\equiv 25 \pmod {42}
\end{align}
Now we have that pesky $2^{2014}\pmod{42}$, where the exponent base isn't coprime to the modulus.  At this point, I'll just say "use the CRT like @GohP.iHan did in his answer," because I wasn't sure how to approach that.  You should have that $2^{2014} \equiv 16$.
Multiplying, we have an answer: $$2014^{2014} \equiv 16\cdot 37\cdot 25 = 14800 \equiv 16 \pmod 42$$
So, the final answer is $3^{16}\equiv 25\pmod {98} $
