[proofplan] The proof uses two standard facts about the [characteristic polynomial](/page/Characteristic%20Polynomial). First, the constant term of $\chi_A(t)$ is $(-1)^n\det A$, so invertibility of $A$ implies that this coefficient is nonzero. Second, the [Cayley-Hamilton theorem](/theorems/865) says that substituting $A$ into its own characteristic polynomial gives the zero matrix. After writing this matrix identity, we multiply by $A^{-1}$ and solve for $A^{-1}$. [/proofplan] [step:Identify the constant term as a nonzero scalar] Let $\det: M_n(k)\to k$ denote the determinant map. By [citetheorem:7910], the constant coefficient of the characteristic polynomial satisfies \begin{align*} c_0=(-1)^n\det A. \end{align*} Since $A$ is invertible, $\det A\ne 0$ in $k$. Also $(-1)^n\ne 0$ in the field $k$, because it is either $1$ or $-1$, and in characteristic $2$ these are equal to the nonzero element $1$. Therefore \begin{align*} c_0\ne 0. \end{align*} Thus $c_0^{-1}\in k$ is defined. [guided] The first issue is to justify that the scalar $c_0^{-1}$ appearing in the formula exists. Let $\det: M_n(k)\to k$ denote the determinant map. By [citetheorem:7910], if \begin{align*} \chi_A(t)=t^n+c_{n-1}t^{n-1}+\cdots+c_1t+c_0, \end{align*} then the constant coefficient is \begin{align*} c_0=(-1)^n\det A. \end{align*} The hypothesis that $A$ is invertible means precisely that its determinant is nonzero: \begin{align*} \det A\ne 0. \end{align*} The factor $(-1)^n$ is a nonzero element of $k$: it is either $1$ or $-1$, and even in characteristic $2$ the element $-1$ equals $1$, not $0$. Multiplying a nonzero element of a field by another nonzero element gives a nonzero element, so \begin{align*} c_0=(-1)^n\det A\ne 0. \end{align*} Hence $c_0$ has a multiplicative inverse $c_0^{-1}\in k$. [/guided] [/step] [step:Apply Cayley-Hamilton to obtain a matrix identity] Set $c_n:=1$. The polynomial expansion can be written as \begin{align*} \chi_A(t)=\sum_{j=0}^n c_jt^j. \end{align*} By the [Cayley-Hamilton theorem](/theorems/923), which applies because $A$ is an $n\times n$ matrix over the field $k$, substituting the matrix $A$ into its characteristic polynomial gives the zero matrix: \begin{align*} \chi_A(A)=0. \end{align*} Therefore, where $0$ denotes the zero matrix in $M_n(k)$, \begin{align*} \sum_{j=0}^n c_jA^j=0. \end{align*} Separating the constant term gives \begin{align*} \sum_{j=1}^n c_jA^j+c_0I_n=0. \end{align*} [/step] [step:Multiply by the inverse and solve for $A^{-1}$] Since $A$ is invertible, left multiplication by $A^{-1}$ is valid in the matrix ring $M_n(k)$. Multiplying \begin{align*} \sum_{j=1}^n c_jA^j+c_0I_n=0 \end{align*} on the left by $A^{-1}$ gives \begin{align*} A^{-1}\left(\sum_{j=1}^n c_jA^j+c_0I_n\right)=0. \end{align*} Using distributivity, associativity of matrix multiplication, and $A^{-1}A^j=A^{j-1}$ for every $j\in\{1,\dots,n\}$, this becomes \begin{align*} \sum_{j=1}^n c_jA^{j-1}+c_0A^{-1}=0. \end{align*} Solving for $A^{-1}$ and using $c_0\ne 0$ yields \begin{align*} A^{-1}=-c_0^{-1}\sum_{j=1}^n c_jA^{j-1}. \end{align*} Since $c_n=1$, this is exactly \begin{align*} A^{-1}=-c_0^{-1}(A^{n-1}+c_{n-1}A^{n-2}+\cdots+c_2A+c_1I_n). \end{align*} This proves the claimed polynomial expression for $A^{-1}$. [/step]