Finding the last digit of $3^{729}$ I'm practicing for my algebra exam but I stumbled on a question I don't know how to solve. 

Let $N = 3^{729}$. What is the last digit of $N$?

The example answer says 

Since $\gcd(3, 10) = 1$, check that
  $3^4 = 81 = 1 \pmod {10}$:
  Now, $729 = 182 \times 4 + 1,$ so we get we get (that might be a typo or missed a step)
  $3^{729} = 3 \pmod {10}$. 

Can anybody help me with this question? Thanks so much!

$3^0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 1\equiv3^0\mod 10$
$3^1\equiv 3\times 1\equiv 3\equiv3^1\mod 10$
$3^2\equiv 3\times3\equiv 9\equiv3^2\mod 10$
$3^3\equiv 3\times9\equiv 7\equiv3^3\mod 10$
$3^4\equiv 3\times7\equiv 1\equiv3^0\mod 10$
$3^5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^1\mod 10$
$3^6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^2\mod 10$
$3^7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^3\mod 10$
$...$
$3^{728}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^0\mod 10$
$3^{729}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^1\mod 10$
