= Solution
The <Cayley-Hamilton theorem> states that every square complex <matrix> satisfies its own <characteristic polynomial>: if $\chi_A(t)=\det(tI-A)=t^n+c_{n-1}t^{n-1}+\cdots+c_0$, then
$$
\boxed{A^n+c_{n-1}A^{n-1}+\cdots+c_0I=0.}
$$
For a proof, expand the <adjugate matrix> as $\operatorname{adj}(tI-A)=\sum_{k=0}^{n-1}B_kt^k$. The <adjugate identity> gives
$$
(tI-A)\sum_{k=0}^{n-1}B_kt^k=\chi_A(t)I.
$$
Comparing coefficients yields $-AB_0=c_0I$, $B_{k-1}-AB_k=c_kI$ for $1\le k\le n-1$, and $B_{n-1}=I$. Multiply the middle identities on the left by $A^k$, include the first identity, and add $A^nB_{n-1}=A^n$. The terms involving the $B_k$ cancel consecutively:
$$
-AB_0+\sum_{k=1}^{n-1}(A^kB_{k-1}-A^{k+1}B_k)+A^nB_{n-1}=0.
$$
Their right sides sum to $c_0I+\sum_{k=1}^{n-1}c_kA^k+A^n$. This proves the theorem without any substitution of noncommuting <matrices> into a scalar <polynomial> identity.
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