This proof is extreeeeemely boring, but still I must get it right.
Let $x = x_0 + x_1 i + x_2 j + x_3 k \in \mathbb{H}$ (the Hamilton quaternions). Conjugation is defined as: $$x^\ast = x_0 - x_1 i - x_2 j - x_3 k \, .$$ Show that for any $x, y\in \mathbb{H}$: $$(x y)^\ast = y^\ast x^\ast$$ Proof: We use a different notation, let $$i_0 = 1, \quad i_1 = i, \quad i_2 = j, \quad i_3 = k \, .$$ Then we have $$i_1^2 = i_2^2 = i_3^2 = -1, \quad i_1 i_2 = -i_2 i_1 = i_3, \quad i_2 i_3 = -i_3 i_2 = i_1 \quad \text{and}\quad i_3 i_1 = -i_1 i_3 = i_2 \, .$$ For $r, s \geq 1$ and $r \neq s$ we see: $$ (i_r i_s)^\ast = - (i_r i_s) = i_s i_r = (-i_s) (-i_r ) = i_s^\ast i_r ^\ast \, .$$ This holds in general for all $0 \leq r, s \leq 3$ (the other cases are easily checked). Finally we get:
$$(x y)^\ast = \Biggl(\Bigr(\sum_{r=0}^3 x_r i_r\Bigl) \cdot \Bigr(\sum_{s=0}^3 y_s i_s\Bigl) \Biggr)^\ast =$$ $$= \Biggl(\sum_{r, s = 0}^3 x_r y_s i_r i_s \Biggr)^\ast = \sum_{r, s = 0}^3 x_r y_s (i_r i_s)^\ast = \sum_{r, s = 0}^3 x_r y_s i_s^\ast i_r^\ast =$$ $$ = \Bigr(\sum_{s=0}^3 y_s i_s^\ast \Bigl) \cdot \Bigr(\sum_{r=0}^3 x_r i_r^\ast\Bigl) = y^\ast x^\ast \, .$$ $\square$
Everything ok? Thank you!