= Solution
Put $d=\deg D$, choose a $K$-basis of $D$, and let $N(X_1,\ldots,X_{d^2})$ be its <reduced norm> polynomial. For $a\in K^\times$, consider
$$
F(X_1,\ldots,X_{d^2},T)=N(X_1,\ldots,X_{d^2})-aT^d.
$$
This has degree $d$ and $d^2+1>d^2$ variables. The $C_2$ property gives a nontrivial zero $(x,t)$. Necessarily $t\ne0$: otherwise $N(x)=0$ forces $x=0$ in the <division algebra>, making the whole coordinate vector zero. Homogeneity now gives $N(x/t)=a$. The value zero is achieved by the zero element. Hence
$$
\boxed{\operatorname{Nrd}:D\longrightarrow K\text{ is surjective}}.
$$
This is the <surjectivity of reduced norms over C2 fields>, and the proof also gives the multiplicative surjection $D^\times\to K^\times$. The argument works for degree one as well.
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