What is the remainder when $7^{7^{7^{7..........Infinity }}}$ is divided by $5$? What is the remainder when $7^{7^{7^{7.........Infinity }}}$ is divided by $5$ ?

My try :
$7^7$ when divided by $5$ gives the remainder $3$,and 
similarly, $7^{7^7}$ when divided by $5$ again gives the remainder $3$,and 
so, i know that upto infinity it will give remainder $3$.

But, Above approach does not shows any real Maths. How to approach for such questions ?

Edit : 
Can I stop the power tower to something like $7^{4k + R}$ ?
 A: First off, notice that "$7^{7^{\cdots\infty}}$" is not a number - what does it mean to evaluate an infinite power tower?
What I think you mean is most easily expressed using the operation of tetration. Let $^n7$ denote a length-$n$ power tower of $7$s (so $^37 = 7^{7^7}$). Then the question seems to be "what is the remainder of $^n7$ when divided by $5$, for any $n$?"
My other concern is with your argument - two examples does not a proof make. I could similarly argue "$3$ is prime, $5$ is prime, therefore every odd number is prime". What you want to do is use induction: show that if $^n7$ has remainder $3$ when divided by $5$, then so does $^{n+1}7$. That's actually pretty easy - an answer posted while I was typing this gave a good explanation. But once you have that argument, and the fact that $^27 = 7^7$ works, you now know that $^n7$ has remainder $3$ for every $n > 1$.
A: $7^1\equiv2\pmod5$
$7^2\equiv4\pmod5$
$7^3\equiv3\pmod5$
$7^4\equiv1\pmod5$
Thus,
$7^n=7^{4k+(n\mod4)}=(7^4)^k\times7^{n\mod4}\equiv1^k\times7^{n\mod4}=7^{n\mod4}\pmod5$
Via similar reasoning,
$7^n=7^{2k+(n\mod2)}=(7^2)^k\times7^{n\mod2}\equiv1^k\times7^{n\mod2}=7^{n\mod2}\pmod4$
Finally,

$$7^{7^{7^n}}\equiv7^{(7^{(7^n\mod2)}\mod4)}=7^3\equiv3\pmod5$$

A: let $a_1=7$ and let $a_{n+1}=7^{a_n}$. It is clear that $a_n\equiv 3 \bmod 4$ because $7$ is congruent to $3\bmod 4$, and we are raising this number to an odd exponent.
Therefore, by fermats theorem we have $7^{a_n}\equiv 7^3\equiv 3 \bmod 5$.
So $a_n\equiv 3\bmod 5$ for $n>1$.
A: $7\equiv 2 \bmod 5$
$2^x \equiv 2^{x \bmod \phi(5)} \bmod 5$
$2^x \equiv 2^{x \bmod 4} \bmod 5$
$7 \equiv 3\bmod 4$
$3^x \equiv -1^{x} \equiv -1^{x\bmod 2} \bmod 4$
$7^{\text{anything}} \equiv 1 \bmod 2$
$7^{7^{\text{anything}}} \equiv 3^{7^{\text{anything}}} \equiv 3  \bmod 4$
$7^{7^{7^{\text{anything}}}} \equiv 2^{3^{7^{\text{anything}}}} \equiv 2^{3} \equiv 3 \bmod 5$
A: ${\rm mod}\,\ \color{#c00}4\!:\,\ \color{#0a0}{7^{\large K}}\!\equiv (-1)^{\large K}\!\equiv\, \color{#c00}{\pm 1},\ $ i.e. $\,+1 $ for even $K;\,$  $\,-1\,$ for odd $K.\,$  Therefore
${\rm mod}\,\ 5\!:\ 7^{\Large\color{#0a0}{7^{\LARGE K}}}\!\!\!\equiv 2^{\large\color{#c00}{\pm1+4N}}\!\equiv 2^{\large \pm1} {\underbrace{(2^{\large 4})}_{\large \equiv\: 1\ }}^{\!\large N}\!\!\equiv 2^{\large \pm1}\!\equiv \pm 2,\ $ i.e. $\, 2\,$ for even $K;\,$ $-2\,$ for odd $K$ (OP) 
