<s>[INST] 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 $7^{4k + R}$ ?
 [/INST] $7\equiv 2 \bmod 5$
$2^x \equiv 2^{x \bmod \phi(5)} \bmod 5$
$2^x \equiv 2^{x \bmod 4} \bmod 5$
$7 \equiv 3\bmod 4$
$3^x \equiv -1^{x} \equiv -1^{x\bmod 2} \bmod 4$
$7^{\text{anything}} \equiv 1 \bmod 2$
$7^{7^{\text{anything}}} \equiv 3^{7^{\text{anything}}} \equiv 3  \bmod 4$
$7^{7^{7^{\text{anything}}}} \equiv 2^{3^{7^{\text{anything}}}} \equiv 2^{3} \equiv 3 \bmod 5$
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