Find the remainder in the following case. Find the remainder when $444^{444^{444}}$ is divided by $7$.
My approach :
$E(7) = 6 $
$444^{444} \pmod 6 = 0$ 
so , $444^0 \pmod 7 = 1$
 A: Yes your approach is indeed correct. I assume by $E(7)$ you mean the Euler totient function $\phi(7)$.
By Euler's theorem/ Fermat's little theorem, we have
$$444^{6} \equiv 1 \pmod{7}$$
Now $6$ divides $444$. Hence, $444^{444} = 6M$. Hence, $$444^{444^{444}} = 444^{6M} = \left( 444^6\right)^M \equiv 1^M \pmod{7} \equiv 1 \pmod{7}$$
A: $a^{\phi(p)}=1(\mod p)$.Here $444^{444}=0(\mod \phi(7)) \implies 444^{444}=k*\phi(7)$ for some integer k.Then, $444^{444^{444}}(\mod 7)= 444^{k*\phi(7)}(\mod 7) = (444^{\phi(7)})^k(\mod 7)=1^k(\mod 7) = 1 $
A: step 1 : Try to bring big no's into a form so that we get remainder of $-1 $ or $1$ , so that it will be easy for us to simplify
$$444 = 4*111 $$ 
        as $ \frac {111}7 $  gives a remainder of $-1$ . So our goal of converting a bigger number to number which gives remainder $1$ or $-1$ is attained 
        and  as $$ -1^{even} = 1$$
so $$ 444^x /7 = (4^x  * 111^x )/7 =   4^x / 7  $$ $$( 111^x \ divided \ by\ 7\ \ \ gives \ remainder \ of \ -1^x\ which\ is\ equal\ to\ 1\ as\ x\ is\ even\ )$$
Where $$ x = 444^{444}  (even) $$ 
So
$$4^x= 4^{444^{444}} = 2^{888^{444}} $$
step 2 : Observe the pattern of the remainders
$$ 2^0 mod 7 = 1 $$
$$ 2^1 mod 7 = 2 $$
$$ 2^2 mod 7 = 4 $$

$$ 2^3 mod 7 = 1 $$
$$ 2^4 mod 7 = 2 $$
$$2^5 mod 7 = 4 $$

so here, for a period of $ 3 $ remainder $ 1 $ repeats 
here  $ 888 $  is a multiple of $3$ , so 
$$  2^{888}\ mod \ 7 =\ 1 $$
$$ 1^{444} \ mod \ 7 = \ 1 $$
Hence Answer is $ 1 $
