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Theorem. (Inverse function theorem) If the Jacobian of a funciton $f: U \to V$ between open neighbourhoods of $0$ in $\mathbb{C}^n$ is non-degenerate then $f$ has a holomorphic inverse.

Propostion. (Griffiths-Harris "Principles of algebraic geometry", p.19). If $f: U \to V$ is a bijective holomorphic map between open neigbourhoods of $0$ in $\mathbb{C}^n$ then $f$ has a holomorphic inverse.

Proof. For n=1, suppose that $J(f)=0$ then after coordinate change $f(z)=z^2 g(z)$, or even $f(z)=z^k g(z)$, where $g(z)$ is non-zero on $U$. Then by possibly shrinking $U$ to a smaller neighbourhood of $0$ we can find a function $h(z)$ such that $h^k(z) = g(z)$ on $U$. Then $f(z)=(z\,h(z))^k$ and therefore $f$ is not bijective. 

Suppose the statement is proved for $n-1$. Let $f$ be bijective and let the Jacobian $J(f)(0)$ be degenerate, i.e. $\mathrm{rk} J(f)(0) = k < n$. Suppose that the coordinate functions $f_1, \ldots, f_n$ are ordered so that $\left(\dfrac{\partial f_i}{\partial z_j}\right)_{i,j=1}^k$ is non-degenerate. Consider the restriction

$f': U \cap f^{-1}(\mathbb{C}^{n-k}) \to V \cap \mathbb{C}^{n-k}$

where $\mathbb{C}^{n-k}$ is the embedding $(z_1, \ldots, z_{n-k}) \mapsto (0, \ldots,0, z_1, \ldots, z_{n-k})$.

Let $g: (z_1, \ldots, z_n) \mapsto (f_1(z), \ldots, f_k(z), z_{k+1}, \dots, z_n)$. Since $J(g)(0)$ is non-degenerate, $g$ is invertible in some neighbourhood of $0$. Then $f'\circ g^{-1}$ maps $\{0, \ldots, 0\} \times U'$  to $\{0, \ldots, 0\} \times V'$, where $U',V' \subset \mathbb{C}^{n-k}$. Call $h$ the restriction of $f' \circ g^{-1}$ to last $n-k$ coordinates.

If $k > 0$ then we can apply the induction hypothesis to $h$. It is bijective, and therefore $J(h)(0)$ is non-degenerate. But $h$ coincides with last $n-k$ coordinates of $f$, and hence $J(h)(0)$ is a $(n-k)\times(n-k)$ minor in $J(f)(0)$. It follows that $J(f)(0)$ is non-degenerate, which contradicts the initial assumption.