[proofplan] We expand $\det(tI_n-A)$ by the Leibniz formula and track exactly which products contribute to the coefficient of $t^{n-r}$. A factor of $t$ can occur only from a diagonal entry, so choosing $n-r$ factors of $t$ forces the permutation to fix the corresponding $n-r$ indices. The remaining $r$ indices form a subset $S$, and the contribution over all permutations of $S$ is precisely $\det(-A_S)=(-1)^r\det A_S$. Summing these contributions over all subsets $S$ of size $r$ gives the desired coefficient formula. [/proofplan] [step:Expand the determinant and isolate the possible sources of powers of $t$] Let $N:=\{1,\dots,n\}$. Let $S_n$ denote the symmetric group of all bijections $\sigma:N\to N$, and let \begin{align*} \operatorname{sgn}:S_n\to\{-1,1\}\subset k \end{align*} denote the [sign homomorphism](/theorems/778) on permutations. For $\sigma \in S_n$, define \begin{align*} m_\sigma(t):=\operatorname{sgn}(\sigma)\prod_{i=1}^{n}\left(\delta_{i,\sigma(i)}t-a_{i,\sigma(i)}\right)\in k[t], \end{align*} where $\delta_{ij}$ is the Kronecker delta. By the [Leibniz formula for the determinant](/theorems/7882), \begin{align*} \chi_A(t)=\sum_{\sigma\in S_n}m_\sigma(t). \end{align*} For a fixed $\sigma\in S_n$, the factor indexed by $i$ contains a $t$-term exactly when $\sigma(i)=i$, because then $\delta_{i,\sigma(i)}=1$; if $\sigma(i)\ne i$, the factor is the constant $-a_{i,\sigma(i)}$. Hence a contribution to the coefficient of $t^{n-r}$ in $m_\sigma(t)$ is obtained by choosing a subset $T\subset N$ with $|T|=n-r$ such that $\sigma(i)=i$ for every $i\in T$, taking the $t$-term from the factors indexed by $T$, and taking the constant term from the remaining factors. Therefore the coefficient of $t^{n-r}$ in $\chi_A(t)$ is \begin{align*} \sum_{\substack{\sigma\in S_n, \; T\subset N, |T|=n-r, \; \sigma(i)=i\text{ for all }i\in T}} \operatorname{sgn}(\sigma)\prod_{i\in N\setminus T}(-a_{i,\sigma(i)}). \end{align*} [/step] [step:Show that each contribution is supported on a principal index set] Let $T\subset N$ satisfy $|T|=n-r$, and define its complement $S:=N\setminus T$. Then $|S|=r$. Suppose $\sigma\in S_n$ fixes every element of $T$. Since $\sigma$ is a bijection from $N$ to $N$ and $\sigma(T)=T$, it follows that $\sigma(S)=S$. Thus the restriction $\sigma|_S:S\to S$ is a permutation of $S$. Conversely, if $S\subset N$ has $|S|=r$ and $\tau:S\to S$ is a permutation, there is a unique permutation $\widetilde{\tau}\in S_n$ defined by \begin{align*} \widetilde{\tau}(i)=i\quad\text{for }i\in N\setminus S \end{align*} and \begin{align*} \widetilde{\tau}(i)=\tau(i)\quad\text{for }i\in S. \end{align*} This gives a one-to-one correspondence between pairs $(\sigma,T)$ contributing to the coefficient of $t^{n-r}$ and pairs $(S,\tau)$ where $S\subset N$, $|S|=r$, and $\tau$ is a permutation of $S$. [guided] The only subtle point in the coefficient extraction is that the choice of $t$-terms is not independent of the permutation. A $t$ can only come from the entry in position $(i,i)$ of $tI_n-A$. In the Leibniz product attached to a permutation $\sigma$, the factor in row $i$ is \begin{align*} \delta_{i,\sigma(i)}t-a_{i,\sigma(i)}. \end{align*} This factor contains a $t$-term exactly when $\sigma(i)=i$. Now fix a set $T\subset N$ of rows from which we choose the $t$-term. If $|T|=n-r$, then these choices would produce a factor $t^{n-r}$, but only provided that $\sigma(i)=i$ for every $i\in T$. Let $S:=N\setminus T$. Since $\sigma$ fixes every element of $T$, we have $\sigma(T)=T$. Because $\sigma:N\to N$ is bijective, the complement must also be preserved: \begin{align*} \sigma(S)=S. \end{align*} Thus the non-$t$ part of the product involves only rows and columns indexed by $S$. Conversely, if $S\subset N$ has $|S|=r$ and $\tau:S\to S$ is any permutation, then extending $\tau$ by the identity on $N\setminus S$ gives a unique permutation $\widetilde{\tau}\in S_n$. For this extended permutation, choosing the $t$-term from every index in $N\setminus S$ and the constant term from every index in $S$ gives a contribution to the coefficient of $t^{n-r}$. This proves that all contributions are exactly parametrized by subsets $S$ of size $r$ together with permutations of $S$. [/guided] [/step] [step:Identify the contribution from each principal submatrix] Fix a subset $S\subset N$ with $|S|=r$. Write $S=\{s_1<\cdots<s_r\}$. Let $S_r$ denote the symmetric group of all bijections $\{1,\dots,r\}\to\{1,\dots,r\}$. For a permutation $\tau:S\to S$, let $\tau_S\in S_r$ be the induced permutation defined by \begin{align*} \tau(s_j)=s_{\tau_S(j)}\quad\text{for each }j\in\{1,\dots,r\}. \end{align*} The extension $\widetilde{\tau}\in S_n$ fixing $N\setminus S$ has the same sign as $\tau_S$, because its permutation matrix is obtained from the permutation matrix of $\tau_S$ by inserting fixed standard basis vectors in the rows and columns indexed by $N\setminus S$. Therefore the total contribution associated with this fixed subset $S$ is \begin{align*} \sum_{\tau:S\to S\text{ bijective}} \operatorname{sgn}(\widetilde{\tau})\prod_{i\in S}(-a_{i,\tau(i)}). \end{align*} Using the increasing order $S=\{s_1<\cdots<s_r\}$, this becomes \begin{align*} \sum_{\pi\in S_r}\operatorname{sgn}(\pi)\prod_{j=1}^{r}(-a_{s_j,s_{\pi(j)}}). \end{align*} By the Leibniz formula for the determinant, this is exactly $\det(-A_S)$. [/step] [step:Convert $\det(-A_S)$ to the stated sign convention and sum over all $S$] For each subset $S\subset N$ with $|S|=r$, the matrix $A_S$ is an $r\times r$ matrix over $k$. Multiplying every row of $A_S$ by $-1$ gives the matrix $-A_S$, so multilinearity of the determinant in the rows gives \begin{align*} \det(-A_S)=(-1)^r\det A_S. \end{align*} Combining the preceding coefficient computation with this identity yields \begin{align*} c_{n-r}=\sum_{\substack{S\subset N, |S|=r}}\det(-A_S). \end{align*} Thus \begin{align*} c_{n-r}=(-1)^r\sum_{\substack{S\subset N, |S|=r}}\det A_S. \end{align*} Since $N=\{1,\dots,n\}$, this is precisely the asserted formula. [/step]