Proving $((a\bmod n) (b\bmod n))\bmod n = ab\bmod n $ 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.
 A: Note that by the Chinese Remainder Theorem, for any $k\gt 0$, and for any integers $a$ and $b$, there exists an integer $x$ such that
$x\bmod k = a\bmod k$ and $x\bmod (k+1) = b\bmod (k+1)$. That is: the remainders of $x$ modulo $k$ and modulo $k+1$ are completely unrelated. So I do not see how you are going to be able to leverage "knowing" the result modulo $k$ into a proof of the result modulo $k+1$, unless you simply prove it directly modulo $k+1$.
So it is really simpler to show that the result holds modulo $n$ directly, for any $n\gt 0$.
Remember that $x\bmod n = r$ if and only if $0\leq r\lt n$ and $x-r$ is a multiple of $n$. 
So first show that $ab - (a\bmod n)(b\bmod n)$ is a multiple of $n$. For example, if $a\bmod n = r$ and $b\bmod n = s$, then $a-r$ and $b-s$ are both multiples of $n$; then $(a-r)b$ is a multiple of $n$, and $r(b-s)$ is a multiple of $n$, so...
A: Let $c = a\bmod n$, true iff $a = jn +c$ for some integer $j$.  Similarly  $d = b\bmod n$ iff $b = kn +d$ for some integer $k$.  
So, 
$ab\bmod n = [(jn+c) \times (kn+d)]  \bmod n $
$= [(jkn+jd+kc)n + c \times d] \bmod n $ 
$ = c \times d \bmod n $ 
$=  [(a\bmod n)\times (b\bmod n)]\bmod n$, 
which is what you wanted to prove.
