How to evaluate $1234^{1234} \pmod{5379}$? Note: $5379 = 3 \times 11 \times 163$.
I tried Chinese Remainder Theorem and Fermat's Little Theorem, got as far as:
$$
1234^{1234} = 1 \pmod{3} \\
1234^{1234} = 5 \pmod{11}
$$
With a bit more work:
$$1234^{1234} = 93^{100} \pmod{163}$$
But $93^{100}$ doesn't really help?
WolframAlpha tells me that $\phi(5379)=3240>1234$
So I can't use Euler's Theorem?
N.B This appeared on a 1st year undergrad problem sheet. So presumably, not too much technology is required. 

$93^{100}$ can be computed with the Fast Exponentiation algorithm. First note $93^2\equiv 10\mod163$. Hence all we have to compute is $10^{50}\mod 163$
It can be done with a hand-held calculator. First note  the exponent in base $2$ is written as $110010$, and we use these digits (from right to left) in the algorithm: at each step, the number $a=10$ is squared and reduced modulo $163$. The power $p$ is initially equal to $1$ and, if the digit in the step is $1$, we multiply the previous value of $p$ by the actual value of $a$ and reduce it modulo $163$: 
$$\begin{array}{c|cr|cr}
\text{exponent}&a&&p\\
\hline
0&a&10&1&1\\
1 & x^2 & 100 & x^2&100\\
0 & x^4 & 57& x^2&100\\
0 & x^8 & -11& x^2&100\\
1 & x^{16} & -42& x^{18}&38\\
1 & x^{32} &-29& x^{50}& 39\\
\end{array}$$
There remains to solve the system of congruences:
$$\begin{cases}
x\equiv 1\mod 3\\x\equiv 5\mod 11\\x\equiv39\mod163
\end{cases}$$
As $4\times 3-11=1$, the solution of the first two congruences is  $x\equiv5\cdot 4\cdot3-1\cdot 11=49\equiv16\mod33$.
Morever, a Bézout's relation between $163$ and $33$ is, by the Extended Euclidean algorithm, $16\cdot163-79\cdot33=1$, whence the solution to the system of congruences
$$\begin{cases}
x\equiv 16\mod 33\\x\equiv39\mod163
\end{cases}\iff x\equiv 16^2\cdot163-39\cdot79\cdot33\equiv4603\mod5379.$$
