Graph of the last digit of $x^n$ - why is it symmetric when $n$ is even, and not when $n$ is odd? I have discovered this fact:
"The graph of the last digit of $x^n$ (where $x$ is positive) is asymmetrical if $n$ is odd, and symmetrical if $n$ is even."
What is the logic behind this?
For example, for $n=2$:

for $n=3$:

 A: Hint: $(10-k)^n$ is $k^n\pmod{10}$ if $n$ is even and $10-k^n\pmod{10}$ if $n$ is odd.
Thus $c:k\mapsto k^n\pmod{10}$ is such that $c(10-k)$ is $c(k)$ in the first case (axial symmetry with respect to the line $k=5$) and $10-c(k)$ in the second case (central symmetry with respect to the point $(5,5)$).
A: Observe that, by the binomial theorem,
$$\begin{align*}
(10-x)^n&=\sum_{k=0}^n\binom{n}{k}10^{n-k}(-x)^k\\\\
&=\underbrace{10^n-\binom{n}{1}10^{n-1}x+\cdots(-1)^{n-1}\binom{n}{n-1}10x^{n-1}}_{\text{divisible by 10}}+(-1)^nx^n\\\\\\
&\equiv (-1)^nx^n\bmod 10.
\end{align*}$$
If $n$ is even, we have $(10-x)^n\equiv x^n\bmod 10$, and if $n$ is odd, we have $(10-x)^n\equiv -x^n\bmod 10$.
(The graph then repeats every ten numbers because $x^n\bmod 10$ depends only on $x\bmod 10$, and the last digit of any number $m$ is equal to its least non-negative residue modulo $10$.)
A: If $n = 2k$ then $x^n = x^{2k} = (x^k)^2$. Regardless of the value of $k$, the last digit of $x^k$ can be $0...9$, and as you experimentally see, for all those digits, the graph of the last digit of $x^2 \forall x$ is symmetrical. You could treat this result involving a power of two as a theroem.
If $n=2k+1$ then $x^{2k+1} = (x^2)x$ . First term of last product produces symmetrical output, then as you change the last digit of the second term, namely $x$, you make it assymetrical.
