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!

$3^1 \equiv 3(\mod 10)$
$3^{2} \equiv -1 (\mod 10) $
$3^3 \equiv 7 (\mod 10)$
$3^{4} \equiv +1 (\mod 10) \implies 3^{4k} \equiv +1 (\mod 10)$
$3^{4k} \cdot 3 \equiv 3 (\mod 10)$
$3^{4k} \cdot 3^2 \equiv -1 (\mod 10)$
$3^{4k} \cdot 3^3 \equiv 7  (\mod 10)$
$729= 4k+1 \implies 3^{729} \equiv 3 (\mod 10)$ 
