Is there another way to prove $(x-n)^2 = (n-x)^2$ Let's say $n$ is $4$. So, I came up with the solution below.


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

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


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

You can argue like this: 
$$(x-n)^2 = ((-1)(n-x))^2 = (-1)^2 (n-x)^2 = (n-x)^2.$$
