Evaluate $6^{433} \pmod {21}$ and a proving question Question 1:
Denote $a \mod b$ as $a \% b$, where $a$ and $b$ are some integers
Evaluate 12^32475 % 21 

The following is what I tried:  
12^32475 % 21 = (12^3 % 21)^10825 % 21 = 6^10825 % 21 = ... = 6^433 % 21

But how to continue ?
Question 2:

I don't understand this: n-1= -1 mod n. Where is this from? And does this line become the next line?  
Context: my class is discrete math for computer science. So far for related topics, it covers modular inverse, (extended) GCD algorithm, Fermat's Little Theorem and Chinese Remainder Theorem (CRT). 
 A: We use the fact that $6^{433} \equiv 0 \mod 3$ and $6^{433} \equiv (-1)^{433} \equiv -1\mod 7 $.
Taking $6^{433} = 3k$ , 
$$3k \equiv -1\mod 7 \implies k \equiv 2\mod 7$$
Hence $6^{433} = 3(\,7\lambda + 2\,) = 21 \lambda + 6$.
So $$6 ^{433} \equiv 6 \mod 21$$
A: Question 1. Use the Chinese remainder theorem: 
you have to compute $12^{32475}\bmod 3\equiv 0^{32475}\bmod 3\equiv 0$ and, using lil' Fermat,
$$12^{32475}\bmod 7\equiv 12^{32475\bmod 6}\equiv 12^3\equiv 4\cdot 12\equiv48\equiv -1\mod 7$$
Now a Bézout's relation between $7$ and $3$ is $\;7-2\cdot 3$, so the solution is
$$0\cdot 7-(-1)\cdot 2\cdot 3=6\mod 21.$$
Added.  On the use of the Chinese remainder theorem: When the moduli $a$ and $b$ are coprime, let $ua+vb=1$ be a Bézout's relation between $a$ and $b$. Then, a system of congruences 
$$\begin{cases}
x\equiv \color{blue}\alpha\mod \color{blue}a,\\x\equiv \color{red}\beta\mod \color{red}b,
\end{cases}$$
has a unique solution modulo $\operatorname{lcm}(a,b)=ab$:
$$x\equiv \color{red}\beta\,\color{blue}{ua}+\color{blue}\alpha\, \color{red}{vb}\mod ab. $$
Question 2:  it is simply that in $n-1$, $n$ is $0$ (mod $n$).
A: It is easier to apply $ $ mod Distributive Law $\,\rm= mDL\,$ vs. $\,\rm CRT\,$ in cases like this, i.e.
$\begin{align}\text{by applying}\ \ \color{#c00}ab\bmod \color{#c00}ac\ \,&\smash[b]{\underset{\textstyle\uparrow}=}\,\  \color{#c00}a(b\bmod c)\, = \text{mDL,}\ \text{ to factor out}\,\ \color{#c00}{a = 3}\\
\ \ \ \ \,(\color{#c00}3\cdot 4)^{\large\color{#0a0}{3+6k}}\bmod \color{#c00}{3}\cdot 7\ \,
&\smash[t]{\overset{\textstyle\downarrow}=}\ \, \color{#c00}3(4\cdot\!\! \underbrace{12^{\large 2+6k}\!\bmod 7}_{\large (-2)^{\Large 2} (-2)^{\Large 6k}\,\equiv\ \ \color{#b8f}{4(1)}\!\!\!\!\!\!\!}\!) =  3(4\cdot \color{#b8f}4\bmod 7) = {3(2)}\end{align}$

Note $\,N = 32475 = 3j\ $ by  $\bmod 3\!:\ N\equiv 3\!+\!(2\!+\!4)\!+\!(7\!+\!5)\equiv 0\,$ by casting out threes.
Thus $\,N = \color{#0a0}{3\!+\!6k},\,$ by $\,N\bmod 6 = 3j\bmod 6 = 3\underbrace{(j\bmod 2)}_{1\ {\rm by\ } N\rm\ odd\ }= 3(1),\, $ again by $\,\rm mDL$.

For $(2)$ by definition $\bmod n\!:\,\ n\!-\!1\equiv -1\iff n\mid (n\!-\!1)-(-1) = n;\ $ it's true $\,n\mid n$
More conceptually: $\ \bmod n\!:\ \color{#c00}{n\equiv 0}\,\Rightarrow\, \color{#c00}n-1\equiv \color{#c00}0-1\equiv -1\, $ by the Congruence Sum Rule
Therefore we conclude that  $ \,(n-1)(n-1)\equiv (-1)(-1)\equiv 1\ $ by the Congruence Product Rule
Or $\ \, \color{#c00}{n- 1\equiv -1}\,\Rightarrow\, (\color{#c00}{n-1})^2\equiv (\color{#c00}{-1})^2\equiv 1\ $ by the Congruence Power Rule
