Unit (last digit) of the number $7^{7^{7}}$ In a conversation on facebook, my theory professor said there know how to drive (last digit) of the number $$7^{7^{7}}$$What would be the method?
For I have the slightest idea ..
Oh, I still have no knowledge of Modular Arithmetic ... Preferably do not use this method so that I can understand.
$$7^{49}$$
 A: This will be quite challenging to do without modular arithmetic, but perhaps I can do it in a way that you will be able to understand easily. 
$7$ is a really neat number in that the last digit of powers of $7$ follow a pattern. Let's calculate the first few powers out:
$7^{0} = 1$
$7^{1} = 7$
$7^{2} = 49$
$7^{3} = 343$
$7^{4} = 2401$
$7^{5} = 16807$
If we continue on in this way, we will quickly find that the last digit of powers of $7$ repeats in the pattern $[1, 7, 9, 3],[1, 7, 9, 3],\ldots$. In other words, the only thing we need to know to find the last digit of a power of $7$ is how far from a multiple of $4$ the exponent is. This notion of "distance from $4$" is precisely determined using modular arithmetic, and it's hard to answer your question without it, but I will do my best to keep things clear.
In modular arithmetic, we say two integers $a$ and $b$ are ${\bf congruent}$ modulo $n$ if their difference is divisible by $n$ - that is, if $a-b = k\cdot n$ for some integer $k$. Modular gives us an easy way of quantifying whether how "close" in the integers a number is to being divisible by another number - that is, taking an integer mod $n$ gives us the remainder when we divide by $n$. 
We know that our last digit repeats every $4$ numbers. So if we can determine what remainder is when we divide $7^7$ by $4$, we're done. In this case, we have to use a clever trick. Note that $7$ is congruent to $-1$ modulo $4$. We note this because $7-(-1) = 8 = 2 \cdot 4$. It turns out that when we care about remainders, we can work with two integers which are congruent just the same. That is, mod $4$, $7^7 = (-1)^7 = -1$. Furthermore, we have that $3$ is congruent to $-1$ mod 4, as $3 - (-1) = 4 = 4 \cdot 1$. Hence, $7^{(7^7)}$ has the same last digit as $7^3$. Thus, the last digit is $3$. 
A: Let $a_n$ be the last digit of $7^n$.  Then we have $a_{4n}=1$, $a_{4n+1}=7$, $a_{4n+2}=9$, and $a_{4n+3}=3$.
So, to determine the last digit of $7^{7^7}$, we need to determine the remainder of $7^7$ upon division by $4$?
Let $b_n$ be the remainder of $7^n$ upon division by $4$.  Then we have $b_{2n}=1$ and $b_{2n+1}=3$.  It follows that $b_7=3$, so that $a_{7^7}=3$.
A: $7\equiv7,7^2=49,7^3=343, 7^4=(49)^2=(50-1)^2=2401\equiv1\pmod{100}$
$$7^7=(8-1)^7\equiv-1\pmod 8\equiv-1\pmod 4\equiv3=4a+3\text{ (say)}$$
Now, $7^{7^7}=7^{4a+3}=(7^4)^a\cdot7^3\equiv 1^a\cdot343\pmod{100}\equiv43\pmod{100}\equiv3\pmod{10}$
As in general $a^{m^n}\ne a^{mn}\implies 7^{7^7}\ne 7^{49}$ as $7^7\ne49$
and  $7^{49}=(7^4)^{12}\cdot7\equiv1\cdot7\pmod{100}$
