Last digits number theory. $7^{9999}$? i have looked at/practiced several methods for solving ex: $7^{9999}$. i have looked at techniques using a)modulas/congruence b) binomial theorem c) totient/congruence d) cyclicity. 
my actual desire would be a start to finish approach using totient/congruence. i have figured out how to work the individual steps but not how to combine them and in what order to be able to solve any "end digits" questions.
 A: Note:
$7^4$ ends in $1$
So,
$7^8$, $7^{12}$, $7^{16}$ all end in $1$.
So, $7^{9996}$ ends in $1$. And $7^3$ ends in $3$. So, the answer is $3$
I have used the fact that
$$
\phi(10) = 4
$$
where $\phi$ is the totient function.

Note: Added in response to OP's comment
If we want the last two digits, we note that $\phi(1000)=400$. So
$$
9999 = 9600 + 399$$
So
$$
7^{9999} \equiv 7^{399} \mod 1000
$$
Since $399$ is 1 less than $400$ we can calculate the answer easily. I will show both the long way and the short way.
Long way which works for any power (not all calculations shown):
We divide by two each time to get
$$
399 = 199+200 \\
199= 99 + 100 \\
88= 49 + 50 \\
49= 24 + 25\\
25 = 12 + 13 \\
13 = 6 + 7 \\
7 = 3+ 4\\
3 = 1 +2 
$$
Now  (all mod 1000)
$$
7^1 = 7,~~~ 7^2 = 49\\
7^3 = 7 \cdot 49 = 343, ~~~7^4 = 49 \cdot 49 = 401\\
7^6 = 343 \times 343 = 649~~~ 7^7 = 343 \cdot 401 = 543
$$ and so on.
You can also find $7^{-1} \mod 1000$ as $143$.
So the last 3 digits are 143
A: Find the last three digits of $7^{9999}$
Solution:
By Euler’s theorem
$(7, 10000)$ are relatively prime therefore $$7 ^ {\phi (1000)} \equiv 1 \pmod {1000}$$
$\phi (1000) = 1000 (1-1/2) (1-1/5)$ {$2$ & $5$ are the only prime divisors of $1000$}$= 400.$
Hence $$7^{400} \equiv 1 \pmod {1000}$$
Also $$(7^{400}) ^{25}\equiv 1^{25} \pmod {1000}$$
Hence $$7^{10000}\equiv 1 \pmod {1000}\tag 1$$
We know
$$\begin{align}
1001 \equiv 1 &\pmod {1000}\\
\implies 1 \equiv 1001 &\pmod {1000}\\
\implies 1\equiv 7 \times 143 &\pmod {1000}\tag 2\\
\end{align}$$
By $1$ and $2$ we get
$$7^{10000}\equiv 7 \times 143 \pmod {1000}$$
$$7^{9999}\equiv 143 \pmod {1000}$$
Hence the last three digits for $7^{9999}$ are $$\bbox[border:2px solid red]{143}$$
A: Note that $7^2=49\equiv 9 \mod 10$, since $9^2=81 \equiv 1 \mod 10$, we have that
$$
7^4\equiv 1 \mod 10
$$
Now notice that $4 \mid 9996$ because $4$ divides $96$ (a number is divisible by $4$ iff its last two digits are divisible by $4$). That leaves us with a $7^3$ remaining which we know that $7^3\equiv 3 \mod 10$.
So that means that 
$$
7^{9999}\equiv 7^3 \equiv 1\cdot 3=3 \mod 10
$$
This means the last digit of $7^{9999}$ is $3$.
A: calculate the congruence class of $999\bmod\phi(10^n)$ (call it $c$). then the last n digits are the last n digits of $7^c$.
Use $\phi(10^n)=4(10)^{n-1}$
For 1 digit: $999\equiv3\bmod4$ so it's $3$
for 2 digits $999\equiv39\bmod 40$ so its $43$
for 3 digits $999\equiv 199 \mod400$ so its $143$
for more than 3 Euler saves you no time. You can try calculating Charmicael's Lambda which is $2^{n-2}\cdot5^{n-1}$for $n\geq3$
