1.) May I seek your advice on how to prove the following assertion(without recourse to (2)):
If $ 2014 \equiv 14 \ (\text{mod} \ 2000), $ then $2014^{2014} \equiv 14^{14} \ (\text{mod 2000}).$
2.) Also, how we prove from Carmichael's theorem that, "If $a \equiv b \ (\text{mod m})$ and $c \equiv d \ (\text{mod} \ \lambda(m) ),$ then $a^c \equiv b^d \ (\text{mod m})$ ", where $\lambda (m)$ is defined as such: http://en.wikipedia.org/wiki/Carmichael_function
Thank you.
First of all,
$\displaystyle2014=2000+14\implies 2014^n=(2000+14)^n=14^n\pmod{2000}$ for any integer $n$
Now, $\displaystyle14^{2014}-14^{14}=14^{14}\left(14^{2000}-1\right)$
As $2000=16\cdot125$ where $(16,125)=1$
Clearly, $16|14^{14}=2^{14}\cdot7^{14}$
Now as $(14,125)=1$ as $\displaystyle\phi(125)=5^2(5-1)=100, 14^{100}\equiv1\pmod{125}$
$\displaystyle\implies 14^{2000}=(14^{100})^{200}\equiv1^{200}\pmod{125}\equiv1 $
$\begin{eqnarray}A&\equiv& a\!\!\pmod{n}\\ B&\equiv& b\!\!\pmod{\phi(n)}\end{eqnarray}\bigg\rbrace \Rightarrow\ A^B\overset{\rm\color{#c00} P}\equiv a^B\overset{\rm\color{#0a0} E}\equiv a^b\pmod n\ $ by $\,\rm\color{#C00}Product\ $ Rule, and $\rm\color{#0a0}Euler's\ \phi\ $ Theorem,
presuming $\,\color{#0a0}{(a,n) = 1}.\,\ $Carmichael's theorem is a sharpening Euler's theorem that arises primarily by combining the moduli using $\rm\,lcm\,$ vs. product. You can find a proof of Carmichael's theorem in Yimin Ge, A note on the Carmichael Function.
$\lambda(n)$ is the smallest number such that for any k we have $k^{\lambda(n)}\equiv 1 \bmod n$ for any $k$. suppose we have $b\equiv c \bmod \lambda n$. Then $b=c+x\lambda(n)$ for some integer x. Then we have $k^{c}=k^{b+x\lambda(n)}=k^b*(k^{\lambda(n)})^x\equiv k^b*1^x\equiv k^b \bmod n$
