I am self studying number theory and have come across a problem involving Fermat's Little Theorem that I cannot seem to solve. The question asks to find the least remainder of $71^{71} \pmod {17}$.
I started by noticing that $71^{71}=71^{(17-1)4} \times 71^{7}$, which, by Fermat's, means my answer is congruent to $71^{7} \mod17$.
Is this the right first step, or is there a better way to solve this problem?
Okay so you are doing it right!
First you do as what OP did reduce $71$ to relate it with Fermat's little theorem
$71^{71} \equiv 71^{4(17-1)}\times 71^7 \mod 17$
Now, $(71^{16})^4 \equiv 1^4 \mod 17$ (Using Fermat's little theorem)
Which means we are left with $71^7 \mod 17$
$71 \equiv 4(17-1) +7\mod 17$
$\implies 71 \equiv 3 \mod 17$
$\implies 71^7 \equiv 3^7 \mod 17$
$\implies 3^{4} \cdot 3^{3} \equiv 81 \times 27 \mod 17$
$\implies 81 \times 27 \equiv (-4)(-7)\mod 17$
$\implies 28 \equiv 11 \mod 17$
