One must show that $$ \overline{[GL_\infty(A),GL_\infty(A)]}= (GL_\infty(A))_0. $$ The proof that the left side is in the right side is the easier one: Let $a$ be in the left side. Then $a=bc$, where $b\in [GL_\infty(A),GL_\infty(A)]$ and $\|c-1\|<1$. Commutators belong to $(GL_\infty(A))_0$, because $K_1(A)$ is abelian (or because of the proof of this fact). So $b\in (GL_\infty(A))_0$. Also, $c=e^h$ for some $h\in M_\infty(A)$, and so it is connected to $1$ by the path $t\mapsto e^{th}$.

Choose now $a\in (GL_\infty(A))_0$. It can be written as finite product of exponentials $e^h$, with $h\in M_\infty(A)$. So it is enough to show that these exponentials belong to the left side. We first prove this for $h\in M_\infty(A)$ of the form $xy-yx$. In this case one has $$ (e^{x/n}e^{y/n}e^{-x/n}e^{-y/n})^{n^2}\to e^{xy-yx}. $$ The elements on the left side are products of commutators so we get the desired result. Finally, notice that any $h\in M_\infty(A)$ is a limit of commutators. For example, assuming that $h\in A$ for simplicity we can express the matrix $$ \begin{pmatrix} 1 & & & \\ & -\frac 1 n & &\\ && -\frac 1 n &\\ &&&\ddots \end{pmatrix} $$ as a commutator in $M_{n+1}(\mathbb C)$ then multiply by $h$.

One must show that $$ \overline{[GL_\infty(A),GL_\infty(A)]}= (GL_\infty(A))_0. $$ The proof that the left side is in the right side is the easier one: Let $a$ be in the left side. Then $a=bc$, where $b\in [GL_\infty(A),GL_\infty(A)]$ and $\|c-1\|<1$. Commutators belong to $(GL_\infty(A))_0$, because $K_1(A)$ is abelian (or because of the proof of this fact). So $b\in (GL_\infty(A))_0$. Also, $c=e^h$ for some $h\in M_\infty(A)$, and so it is connected to $1$ by the path $t\mapsto e^{th}$.

Choose now $a\in (GL_\infty(A))_0$. It can be written as finite product of exponentials $e^h$, with $h\in M_\infty(A)$. So it is enough to show that these exponentials belong to the left side. We first prove this for $h\in M_\infty(A)$ of the form $xy-yx$. In this case one has $$ (e^{x/n}e^{y/n}e^{-x/n}e^{-y/n})^{n^2}\to e^{xy-yx}. $$ The elements on the left side are products of commutators so we get the desired result. Finally, notice that any $h\in M_\infty(A)$ is a limit of commutators. For example, assuming that $h\in A$ for simplicity we can express the matrix $$ \begin{pmatrix} 1 & & & \\ & -\frac 1 n & &\\ && -\frac 1 n &\\ &&&\ddots \end{pmatrix} $$ as a commutator in $M_{n+1}(\mathbb C)$ then multiply by $h\otimes 1_{n+1}$ to get diag$(h,-h/n,\ldots,-h/n)$ as a commutator.

One must show that $$ \overline{[GL_\infty(A),GL_\infty(A)]}= (GL_\infty(A))_0. $$ The proof that the left side is in the right side is the easier one: Let $a$ be in the left side. Then $a=bc$, where $b\in [GL_\infty(A),GL_\infty(A)]$ and $\|c-1\|<1$. Commutators belong to $(GL_\infty(A))_0$, because $K_1(A)$ is abelian (or because of the proof of this fact). So $b\in (GL_\infty(A))_0$. Also, $c=e^h$ for some $h\in M_\infty(A)$, and so it is connected to $1$ by the path $t\mapsto e^{th}$.

Choose now $a\in (GL_\infty(A))_0$. It can be written as finite product of exponentials $e^h$, with $h\in M_\infty(A)$. So it is enough to show that these exponentials belong to the left side. We first prove this for $h\in M_\infty(A)$ of the form $xy-yx$. In this case one has $$ (e^{x/n}e^{y/n}e^{-x/n}e^{-y/n})^{n^2}\to e^{xy-yx}. $$ The elements on the left side are products of commutators so we get the desired result. Finally, notice that any $h\in M_\infty(A)$ is a limit of commutators. For example, assuming that $h\in A$ for simplicity we can express the matrix $$ \begin{pmatrix} 1 & & & \\ & -\frac 1 n & &\\ && -\frac 1 n &\\ &&&\ddots \end{pmatrix} $$ as a commutator in $M_n(\mathbb C)$ then multiply by $h$.

One must show that $$ \overline{[GL_\infty(A),GL_\infty(A)]}= (GL_\infty(A))_0. $$ The proof that the left side is in the right side is the easier one: Let $a$ be in the left side. Then $a=bc$, where $b\in [GL_\infty(A),GL_\infty(A)]$ and $\|c-1\|<1$. Commutators belong to $(GL_\infty(A))_0$, because $K_1(A)$ is abelian (or because of the proof of this fact). So $b\in (GL_\infty(A))_0$. Also, $c=e^h$ for some $h\in M_\infty(A)$, and so it is connected to $1$ by the path $t\mapsto e^{th}$.

Choose now $a\in (GL_\infty(A))_0$. It can be written as finite product of exponentials $e^h$, with $h\in M_\infty(A)$. So it is enough to show that these exponentials belong to the left side. We first prove this for $h\in M_\infty(A)$ of the form $xy-yx$. In this case one has $$ (e^{x/n}e^{y/n}e^{-x/n}e^{-y/n})^{n^2}\to e^{xy-yx}. $$ The elements on the left side are products of commutators so we get the desired result. Finally, notice that any $h\in M_\infty(A)$ is a limit of commutators. For example, assuming that $h\in A$ for simplicity we can express the matrix $$ \begin{pmatrix} 1 & & & \\ & -\frac 1 n & &\\ && -\frac 1 n &\\ &&&\ddots \end{pmatrix} $$ as a commutator in $M_{n+1}(\mathbb C)$ then multiply by $h$.