How to algebraically arrive at the conclusion that $(5-1)^{11} \cdot 2 = 2*5 (\cdots) + (-1)^{11}$? This week at uni we was doing modular arithmetic in our math course. In an example our teacher showed us during a lecture, one wanted to find out if the sum of two numbers was dividable by five. 
One of the steps was to write the numbers so that you clearly saw that they were dividable by five. A part of one such example is when you write that
$$(5-1)^{11} \cdot 2 = 2*5 (\cdots) + (-1)^{11}.$$
My question though is, how did he arrive there? I understand the logic, but I do not understand how he arrived there. 

How to algebraically arrive at the conclusion that $(5-1)^{11} \cdot 2 = 2*5 (\cdots) + (-1)^{11}$?

 A: The point of modular arithmetic is to be able to adjust your number system, so that you identify a given number (the modulus) with 'zero'. In this case, you identify 5 with 0, so that $(5 - 1)^{11}$ is just $(0 - 1)^{11} \equiv -1 \pmod{5}$. 
To see things 'algebraically', you can also apply the binomial theorem to the left hand side. This gives
$\displaystyle (5-1)^{11} = \sum_{j=0}^{11} \binom{11}{j} 5^j (-1)^{11-j}.$
Clearly, for $j \geq 1$ the term $\binom{11}{j} 5^j (-1)^{11-j}$ is divisible by 5. It remains to look at $j=0$, where the term is just $\binom{11}{0} 5^0 (-1)^{11 - 0} = -1$. 
A: The Binomial Theorem states that, where n is a positive integer:
$$(a + b)^n= a^n + (^nC_1)a^{n-1}b + (^nC_2)a^{n-2}b^2 +\cdots+ (^nC_{n-1})ab^{n-1} + b^n$$
Now put $a=5$ and $b=-1$ and you will get your result as,
$$(5 -1)^{11}= 5^{11} + (^{11}C_1)5^{10}{(-1)} + (^{11}C_2)5^{9}{(-1)}^2 +\cdots+ (^{11}C_{10})5{-1}^{10} + {(-1)}^{11}$$
A: $$(a-b)^n=\binom{n}{0}a^n-\binom{n}{1}a^{n-1}b+\cdots +(-1)^n\binom{n}{n}b^n$$where $\binom{n}{k}=\frac{n!}{(n-k)!k!}$. Here you can get a proof of this result. Now, put $a=5,b=1$, and you will get the result. 
