Is there another way to prove $(x-n)^2 = (n-x)^2$

Question

Let's say $n$ is $4$. So, I came up with the solution below.

1. $(x-4)^2 = (x-4)(x-4) = x^2 - 8x + 16$
2. $(4-x)^2 = (4-x)(4-x) = 16 - 8x + x^2 = x^2 - 8x + 16$

I was wondering if there is another way to proof that $(x-n)^2$ equals to $(n-x)^2$ ?

Answer 1

You can argue like this:

$$(x-n)^2 = ((-1)(n-x))^2 = (-1)^2 (n-x)^2 = (n-x)^2.$$

Answer 2

When $y$ is a real number it is always true that

$$y^2 = (-y)^2 $$