Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$

The proof you've suggested, and the other answers have corrected, is the purely arithmetic way to see the desired equality. The result is fairly clear to see intuitively as well, just ponder the following statement:

"To form a collection of $k$ objects from $n$ total objects, one may choose which $k$ objects to include in the collection, or equivalently, one may choose which $n-k$ objects to exclude from the collection."

You simply failed to distribute the negative sign (i.e., multiplying the difference by $-1$: $$\begin{align} n - (n - k) &= n + -1(n -k) \\ \\& = n + - 1\cdot n - (-1)\cdot k \\ \\ & = n - n + k \\ \\ &= k\end{align}$$

(I.e., Negation distributes over sums and differences.)

$${n\choose n-k}=\frac{n!}{\color{red}{(n-k)}!(n-\color{red}{(n-k)})!} =\frac{n!}{(n-k)!(n - n + k))!} =\frac{n!}{(n-k)!(k)!}= \frac{n!}{k!(n-k)!}$$

$ n - (n-k) = k $, not $ -k $.