[proofplan]
Use the Artin-Wedderburn form of the central simple algebra to replace $A$ by the matrix algebra $M_r(D)$. Morita invariance identifies the low algebraic $K$-groups of $M_r(D)$ with those of the division ring $D$. The standard division-ring computations give $K_0(D)\cong \mathbb Z$ and identify $K_1(D)$ with the abelianization of $D^\times$, equivalently $D^\times/[D^\times,D^\times]$.
[/proofplan]
[step:Replace the central simple algebra by its Artin-Wedderburn matrix algebra]
By the chosen Artin-Wedderburn decomposition, there is an $F$-algebra isomorphism
\begin{align*}
\alpha:A \longrightarrow M_r(D).
\end{align*}
Algebraic $K$-theory is invariant under ring isomorphism, so $\alpha$ induces group isomorphisms
\begin{align*}
K_0(\alpha):K_0(A) \longrightarrow K_0(M_r(D))
\end{align*}
and
\begin{align*}
K_1(\alpha):K_1(A) \longrightarrow K_1(M_r(D)).
\end{align*}
Thus it is enough to compute $K_0(M_r(D))$ and $K_1(M_r(D))$.
[/step]
[step:Apply Morita invariance to reduce from $M_r(D)$ to $D$]
Since $D$ is a unital ring and $r \ge 1$, [Morita Invariance For K0 And K1]([citetheorem:8679]) applies to the matrix ring $M_r(D)$. It gives natural isomorphisms
\begin{align*}
K_i(M_r(D)) \cong K_i(D)
\end{align*}
for $i=0,1$. Therefore
\begin{align*}
K_0(A) \cong K_0(D)
\end{align*}
and
\begin{align*}
K_1(A) \cong K_1(D).
\end{align*}
[guided]
The point of Morita invariance is that algebraic $K_0$ and $K_1$ do not distinguish a unital ring from a full matrix algebra over it. Here the ring to which we apply the theorem is the division ring $D$, and the matrix size is the integer $r \ge 1$ from the Artin-Wedderburn decomposition. These are exactly the hypotheses required by [Morita Invariance For K0 And K1]([citetheorem:8679]).
The theorem gives, for each $i \in \{0,1\}$, an isomorphism
\begin{align*}
K_i(M_r(D)) \cong K_i(D).
\end{align*}
Combining this with the isomorphisms induced by
\begin{align*}
\alpha:A \longrightarrow M_r(D)
\end{align*}
from the previous step gives
\begin{align*}
K_0(A) \cong K_0(M_r(D)) \cong K_0(D)
\end{align*}
and
\begin{align*}
K_1(A) \cong K_1(M_r(D)) \cong K_1(D).
\end{align*}
This reduces the theorem to the two standard computations for a division ring.
[/guided]
[/step]
[step:Compute $K_0$ of the division ring]
Since $D$ is a division ring, [K0 Of A Division Ring]([citetheorem:8677]) gives an isomorphism
\begin{align*}
K_0(D) \cong \mathbb Z.
\end{align*}
Under this isomorphism, the class of a finitely generated projective left $D$-module $P$ maps to its dimension $\dim_D P$. Hence
\begin{align*}
K_0(A) \cong \mathbb Z.
\end{align*}
[/step]
[step:Compute $K_1$ of the division ring using the Dieudonne determinant]
Since $D$ is a division ring, [K1 Of A Division Ring]([citetheorem:8678]) identifies the first algebraic $K$-group of $D$ with the abelianization of its unit group:
\begin{align*}
K_1(D) \cong D^\times/[D^\times,D^\times].
\end{align*}
Here $D^\times$ is the multiplicative group of nonzero elements of $D$, and $[D^\times,D^\times]$ is its commutator subgroup. Combining this with the Morita reduction gives
\begin{align*}
K_1(A) \cong D^\times/[D^\times,D^\times].
\end{align*}
[/step]
[step:Compose the isomorphisms to obtain the stated result]
The preceding steps give chains of group isomorphisms
\begin{align*}
K_0(A) \cong K_0(M_r(D)) \cong K_0(D) \cong \mathbb Z
\end{align*}
and
\begin{align*}
K_1(A) \cong K_1(M_r(D)) \cong K_1(D) \cong D^\times/[D^\times,D^\times].
\end{align*}
These are precisely the two asserted computations of the algebraic $K$-groups of the central simple algebra $A$.
[/step]
