How to prove that a tetration holds the same last digit after a certain rank? Let's say $u_{n} = 7^{u_{n-1}}$ and $u_{0} = 7$, how can we prove that for $n > 0$, that we will get the same last digit?
I know that the last digit of $7^7$ is 3, we can prove it using modulo/congruence. But how can we prove that even if we take this number as a exponent, the last digit will be 3 anyway.
Any hints? Thanks!

$$u_n=7^{{{7^7}^7}^{...}}$$now : we know
$$a^{4P+r}\equiv ^{mod 10}a^r\\$$where r=1,2,3,4$$7\equiv ^{mod 4}3\equiv ^{mod 4} -1\\7^7\equiv ^{mod 4}(-1)^7\equiv ^{mod 4}(-1)\\7^{7^7}\equiv ^{mod 4}(-1)^7\equiv ^{mod 4}(-1)$$and so on...$$7^{{{7^7}^7}^{...}}\equiv ^{mod 4}(-1)\equiv ^{mod 4}(+3) \rightarrow  7^{{{7^7}^7}^{...}}=4p+3 $$now we have :$$u_n=7^{{{7^7}^7}^{...}}\equiv ^{mod 10}7^{4p+3} \equiv ^{mod 10}7^3\equiv ^{mod 10}3\\$$so the last digit of $u_n$ is 3 
