Why is the last digit of $n^5$ equal to the last digit of $n$? I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.
 A: Alternatively, you could prove that $n^5-n$ is divisible by $10$ by induction. If $n=0$, it is obviously true.
Assume true for $n$, we need to show that:
$$(n+1)^5-(n+1) = n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 - (n+1) \\=n^5 - n + 10(n^3+n^2) +5n(n^3+1)$$
is divisible by 10.
But $n^5-n$ is divisibly by $10$ by induction, and $10(n^3+n^2)$ is obviously divisible by $10$, so all you need to show is that $5n(n^3+1)$ is divisible by $10$, which is the same as proving that $n(n^3+1)$ is divisible by $2$.
The fundamental reason for this, as everybody has noted, is due to rules of modular arithmetic.
A: If $\gcd(a, n) = 1$ then by Euler's theorem,
$$a^{\varphi(n)} \equiv 1 \pmod{n}$$
From the tables and as @SeanEberhard stated, $$ \varphi(10) = \varphi(5*2) = 10\left( 1 - \frac{1}{5} \right) \cdot \left(1 - \frac{1}{2} \right)$$
$$= 10\left(\frac{4}{5} \right) \cdot \left(\frac{1}{2} \right) = 4$$
Let $n=10$ and thus,
$$a^{\varphi(10)} \equiv 1 \pmod{10} \implies a^{4} \equiv 1 \pmod{10}$$
Multiply both sides by $a$,
$$a^{5} \equiv a \pmod{10}$$
A: The claim is clearly equivalent to the claim that $n^5 - n$ is divisible by $10$.
By Fermat's little theorem, $n^5 - n$ is divisible by $5$.
$n^5 - n = n(n^4 - 1)$ is even because if $n$ is odd, then $n^4-1$ is even.
Thus, $n^5 - n$ is divisible by both $5$ and $2$, and, these being coprime, it follows that $n^5 - n$ is divisible by $10$.
A: Note, the last digit of $n^5$ can only be affected by the last digit of $n$.  That is, the 1s digit of any power of $n$ will only be affected by the ones digit of $n$, and not the 10s, 100s, or any other digit.  So, you only have 10 cases, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.  All you have to do is try them separately.  $0^5 = 0$,  $1^5 = 1$, $2^5 = 32$, $3^5 = 243$, ..., $9^5 = 59049$.
If you've heard of modular arithmetic, all numbers mod 10 are going to be in the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.  And, this is why we only need consider these 10 cases.
A: $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=n(n^2-1)(n^2-4)+5n(n^2-1)$$
$$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers  }}+5\cdot \underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers  }}$$
Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer
So, $(n-2)(n-1)n(n+1)(n+2)$ is divisible by $5!=120$ and $(n-1)n(n+1)$ is divisible by $3!=6$
$$\implies n^5-n\equiv0\pmod{30}\equiv0\pmod{10}$$
A: I won't explain you that it's true because you discovered it was true !
Fermat's little theorem tells us that :
$$k^5\equiv k\pmod{5}$$
Further $k^5$ will be even or odd when $k$ is even or odd.
You may use the chinese remainder theorem to conclude that indeed : $$k^5\equiv k\pmod{10}$$ 
(this works because of the factorization $10=2\cdot 5$ and that is why other small primes won't share this property...)
We may produce this table of powers with that property :
$$1, 5, 9, 13, 17, 21\cdots$$
The pattern is clear : you'll get the same result for $p=4n+1$ with $n\in \mathbb{N}$.
This is natural since mulplying each digit $k$ by $k^4$ will give $k$ again. 
A: Numbers whose last digits of powers occur in cycles of 4:
21 = 2
22 = 4
23 = 8
24 = 6
25 = 2
31 = 3
32 = 9
33 = 7
34 = 1
35 = 3
71 = 7
72 = 9
73 = 3
74 = 1
75 = 7
81 = 8
82 = 4
83 = 2
84 = 6
85 = 8
Alternating last digits
41 = 4
42 = 6
43 = 4
91 = 9
92 = 1
93 = 9
Powers of numbers ending in 1,5,6 will always end in 1,5,6.
Therefore we can observe from the above cases that,
n^5 and n will always have the same last digit.
