What is the last digit of $.....\Biggl(\biggl(\Bigl(\bigl((7)^7\bigr)^7\Bigr)^7\biggr)^7\Biggr)^7 ......$ $2005$ $7's$ are used in the following number ,so what is its last digit   :
Number: $.....\Biggl(\biggl(\Bigl(\bigl((7)^7\bigr)^7\Bigr)^7\biggr)^7\Biggr)^7 ......$
As i mentioned in the introduction , this number is written using $2005$ times $7's$ and i want to find its last digit.
My work: Its last digit is wanted so i used $\mod(10)$.
After that i formed the expression into $\bigl((7)^{7^{2004}}\bigr)$ , then i decided to use Euler's totient function.
${7^{\phi(10)=4}\equiv 1 \ (mod10)}$
In the next step , i said that ${7^{4k}\equiv 1 \ (mod10)}$ and  i checked over $7^{2004}(mod4)$.Then ,I found that $7^{2004}(mod4)=1$
As a result ,  $\bigl((7)^{7^{2004}}\bigr) \equiv 7^1 (mod 10)=7$
My first question is that is my solution correct ?
Secondly , if my solution is not correct , can you help me to write the correct ?
Thirdly , if my solution is correct but it is long ,can you share any shortcut or beneficial information to make it elegant.
$EDIT$: I want to find its last digit and it is an exponential tower where $n=2005$
 A: $7^{2004}=\big(7^2\big)^{1002}=49^{1002}\equiv1^{1002}=1\pmod 4$
As $\phi(10)=4$,
$7^{7^{2004}}\equiv7^1=7\pmod{10}$
Your solution is correct.
A: If this is indeed a tetration, rather than a repeated taking the seventh power, the answer is different.
Tetration: $7^{\large 7^{\large 7^{\dots}}}$
Repeated power: $((7^7)^7)^7\cdots$
For tetration, the modulus of even a short tower of exponents rapidly settles down to a single value, unaffected by extending the tower, under the influence of the reducing cycles of higher exponents.
Here $7^k \equiv 7^\ell \bmod 10$ when $k \equiv \ell \bmod 4$ since $\lambda(10) = 4$
Then $7^{7^m} \equiv 7^{3^m} \equiv 7^{-1^m} \bmod 10$, directly giving a $2$-cycle, and since we can see that in the given case (of tetration) that $m$ is odd, we have
$$ 7^{\large 7^{\large 7^{\large 7^{\dots}}}} \equiv 7^{\large 7^{\text{[odd #]}}} \equiv 7^{-1} \equiv 3 \bmod 10$$
A: I think the answer is right because I turned up with the same one with this approach:
$7^2\equiv -1\ \left(\text{mod}10\right)$ (Here -1 corresponds to 9)
$7^6\equiv \left(-1\right)^3=-1\ \left(\text{mod}10\right)$
$7^7\equiv 3\ \left(\text{mod}10\right)$
i.e.$7^7\equiv -7\ \left(\text{mod}10\right)$
similarly continuing, we get
$7^{7^7}\equiv \left(-7\right)^7\equiv \left(-7\right)\cdot \left(-7\right)^6\equiv -7\cdot \left(7\right)^6\equiv -7\cdot -1\equiv 7\ \left(\text{mod}10\right)$
and as 2004 is a multiple of 7, all will get cancelled to leave the remainder of 7.
Hence  the last digit is 7.
