I'm trying to show that $a^{|D|} = a$ for every $a\in D$, where $D$ is a finite division ring. My thoughts were to prove that the followign map \begin{equation} \phi : D \to D, \quad \phi(a) = a^{|D|} \end{equation} is bijective, but when I try to prove that $ker\phi = \{0 \}$ i see that i cannot have nilpotents elements on $D$, so I believe I'm not on the right track. So any help would be appreciated.
Thank you in advance
As @Melody pointed out, the multiplicative group of the ring under multiplication has cardinality $\left|D\right|-1$, and thus by Lagrange's theorem, $a^{|D|-1}=1$. So $a^{|D|}= 1 \cdot a= a$.