I am trying to prove a property of modular arithmetic, namely: $[(a\bmod n)\times (b\bmod n)]\bmod n = ab\bmod n.$ I have the basis and hypothesis steps down, but I am having trouble with the hypothesis step:
Proof Let $P(n)$ be the predicate $P(n): [(a \bmod n) \times (b \bmod n)] \bmod n = a b \bmod n.$
Basis step: $$\begin{align*} \ [(a \bmod 1) \times (b \bmod 1)] \bmod 1 &= (a b) \bmod 1\\ \ [( 0 )\times ( 0 )] \bmod 1 &= (a b) \bmod 1 &&\text{(any \number is }0\text{ modulo }1\text{)}\\ 0 \bmod 1 &= (a b) \bmod 1\\ 0 &= 0 &&\text{(true for }n=1\text{)} \end{align*}$$
Hypothesis step
(Assume that $P(n)$ is true for some $n=k$): $[(a \bmod k) \times (b \bmod k)] \bmod k = (a b) \bmod k$
Induction Step
(Prove that $P(n)$ is true for some $n = k + 1$) $\begin{align*} \ [(a \bmod (k+1)) \times (b \bmod (k+1))] \bmod (k+1) &= (a b) \bmod (k+1)\\ \ [(a \bmod k) + 1 \times (b \bmod k) + 1] \bmod (k+1) &= (a b) \bmod (k+1)\\ \end{align*}$
I get to here then I can't figure out how to cancel out the 1's on the left hand side.
Any help would be appreciated.