What will be the units digit of $7777^{8888}$? What will be the units digit of $7777$ raised to the power of $8888$ ? 
Can someone do the math with explaining the fact "units digit of $7777$ raised to the power of $8888$"?
 A: First, you only need the units digit of $7^{8888}$.  The really simple thing is to look for the pattern:  $7^0 \equiv 1, 7^1 \equiv 7, 7^2 \equiv 9, 7^3 \equiv 3, 7^4 \equiv 1 \pmod {10}$ and the pattern repeats.  Since $8888$ is a multiple of $4$, $7777^{8888}$ ends in $1$.
To be a bit more formal, now we have $7777^{8888} \equiv 7^{8888} \equiv (7^4)^{2222}\equiv 1^{2222}=1 \pmod {10}$
A: Note that any integer $n$ can be written as $10a + b$ where $a$ and $b$ are integers and $b$ is between $0$ and $9$ (inclusive). Written in this way, $b$ is the units digit of $n$. For example $7777 = 10(777) + 7$; here $b = 7$.
Instead of raising a number to such a high power ($8888$), let's start with a smaller power, $2$. If we have written our number $n$ in the form $10a + b$, then we have $$n^2 = (10a + b)^2 = 100a^2 + 20ab + b^2.$$ But this is a positive integer and can be written in the aforementioned way. If $b^2 = 10c + d$ is it's representation, then $10(10a^2 + 2b + c) + d$ is the representation of $n^2$. Note that both $b^2$ and $n^2$ have the same units digit, $d$. We can summarise this by the following:

The last digit of $n^2$ is the same as the last digit of $(\text{the last digit of}\ n)^2$.

Using our number as an example, we can check this. Note that $$7777^2 = 60481729$$ which has last digit $9$. The last digit of $7777$ is $7$ and $7^2 = 49$, which also has last digit $9$.
Similar reasoning to the above (in particular the binomial theorem) can be used to prove the following generalisation, where $k$ is any positive integer:

The last digit of $n^k$ is the same as the last digit of $(\text{the last digit of}\ n)^k$.

So for the problem at hand, the above says that the last digit of $7777^{8888}$ is the same as the last digit of $7^{8888}$. The latter is still a huge number, but a little easier to deal with.
As I have explained in a previous answer, you can figure out what the last digit of $7^{8888}$ is in a fairly easy way. In the linked answer, the question is about the last digit of $2^{2006}$, but the process is the same, just with different numbers.
A: the units digit of a number is the same as the number mod 10
so we just need to compute $7777^{8888} \pmod {10}$
first $$7777^{8888} \equiv 7^{8888} \pmod {10}$$ and secondly by Eulers totient theorem $$7^{8888} \equiv 7^{0} \equiv 1 \pmod {10}$$ by Eulers totient theorem (since $\varphi(10)=4$ and $4 \mid 8888$)
so there's a quick way to see that the last digit is a 1
A: $$7777^{1} \equiv 7 \pmod{10}$$
$$7777^{2} \equiv 9 \pmod{10}$$ 
$$7777^{3} \equiv 3 \pmod{10}$$
$$7777^{4} \equiv 1 \pmod{10}$$
$$7777^{5} \equiv 7 \pmod{10}$$
And the relation continues, in general you see that:
$$7777^{4n + 1} \equiv 7 \pmod{10}$$
And
$$7777^{4n} \equiv 1 \pmod{10}$$
$$\frac{8888}{4} = 2222$$
Hence, $7777^{8888} \equiv 1 \pmod{10}$.
