Compute the unit digit of ${{43}^{43}}^{43}$ I am trying to find the unit digit of ${{43}^{43}}^{43}$. Normally I would try to break up the exponent to make it easier though 43 is prime so I can't do that. So instead I first considered $43^{43}$ and tried finding a pattern for the unit digits by looking at the first few values:

$43^0=1$

$43^1=43$

$43^2=1849$

$43^3=79507$

$43^4=3418801$

$43^5=147008443$

...

I then noticed that they repeated every 4 values with a unit digit pattern of {1, 3, 9, 7}.

This is where I am stuck, for a few reasons:


I know I need to prove that this unit digit pattern continues but I'm not quite sure how to do this.
I know the unit digit is 1, 3, 7 or 9, but how do I know which one it is?
After finding the unit digit of $43^{43}$, how do I then use this to find the unit digit of ${{43}^{43}}^{43}$? Lets say $x$ is the unit digit of $43^{43}$ so would I then need to find the unit digit of $43^x$?


Can someone help me with where I am confused, let me know if the proccess I wrote is incorrect, or explain an easier approach to this problem?
$43^{43}\equiv (-1)^{43} \equiv 3 \mod \phi(10) =4$, so $43^{43^{43}}$ has last digit $7$ as number $43^3$.
$43^{43}\equiv (-1)^{43} \equiv 3 \mod \phi(10) =4$, so $43^{43^{43}}$ has last digit $7$ as number $43^3$.