Here is a direct proof along the lines of the standard proof of the Cayley–Hamilton theorem. [This works universally, i.e. over the commutative ring $R=\mathbb{Z}[a_{ij}]$ generated by the entries of a generic matrix $A$.]

The following lemma combining Abel's summation and Bezout's polynomial remainder theorem is immediate.

Lemma Let $A(\lambda)$ and $B(\lambda)$ be matrix polynomials over a (noncommutative) ring $S.$ Then $A(\lambda)B(\lambda)-A(0)B(0)=\lambda q(\lambda)$ for a polynomial $q(\lambda)\in S[\lambda]$ that can be expressed as

$$q(\lambda)=A(\lambda)\frac{B(\lambda)-B(0)}{\lambda}+\frac{A(\lambda)-A(0)}{\lambda}B(0)=A(\lambda)b(\lambda)+a(\lambda)B(0), \quad a(\lambda),b(\lambda)\in S[\lambda] q(\lambda)=A(\lambda)\frac{B(\lambda)-B(0)}{\lambda}+\frac{A(\lambda)-A(0)}{\lambda}B(0)=A(\lambda)b(\lambda)+a(\lambda)B(0) \qquad (*)$$

with $a(\lambda),b(\lambda)\in S[\lambda].$

Let $A(\lambda)=A-\lambda I_n$ and $B(\lambda)=\operatorname{adj} A(\lambda),$A(\lambda)$ [viewed as elements of$S[\lambda]$with$S=M_n(R)$], then$A(\lambda)B(\lambda)=\det A(\lambda)=p_A(\lambda)=p_0+p_1\lambda+\ldots p_n\lambda^nA(\lambda)B(\lambda)=\det A(\lambda)=p_A(\lambda)=p_0+p_1\lambda+\ldots+p_n\lambda^n$$is the characteristic polynomial of A,\ A(0)B(0)=p_0, A and q(\lambda)=p_1+\ldots+p_n\lambda^{n-1}. A(0)B(0)=p_0 \text{ and } q(\lambda)=p_1+\ldots+p_n\lambda^{n-1}$$

Applying $(*),$ we get

$$q(\lambda)=(A-\lambda I)b(\lambda)-\operatorname{adj} A \qquad (**)$$

for some matrix polynomial $b(\lambda).$ b(\lambda)$commuting with$A.$Specializing$\lambda$to$A,$A$ in $(**),$ we conclude that

$$q(A)=-\operatorname{adj} A\qquad \square$$

1

Here is a direct proof along the lines of the standard proof of the Cayley–Hamilton theorem.

The following lemma combining Abel's summation and Bezout's polynomial remainder theorem is immediate.

Lemma Let $A(\lambda)$ and $B(\lambda)$ be matrix polynomials over a (noncommutative) ring $S.$ Then $A(\lambda)B(\lambda)-A(0)B(0)=\lambda q(\lambda)$ for a polynomial $q(\lambda)\in S[\lambda]$ that can be expressed as

$$q(\lambda)=A(\lambda)\frac{B(\lambda)-B(0)}{\lambda}+\frac{A(\lambda)-A(0)}{\lambda}B(0)=A(\lambda)b(\lambda)+a(\lambda)B(0), \quad a(\lambda),b(\lambda)\in S[\lambda] \qquad (*)$$

Let $A(\lambda)=A-\lambda I_n$ and $B(\lambda)=\operatorname{adj} A(\lambda),$ then $A(\lambda)B(\lambda)=\det A(\lambda)=p_A(\lambda)=p_0+p_1\lambda+\ldots p_n\lambda^n$ is the characteristic polynomial of $A,\ A(0)B(0)=p_0,$ and $q(\lambda)=p_1+\ldots+p_n\lambda^{n-1}.$ Applying $(*),$ we get

$$q(\lambda)=(A-\lambda I)b(\lambda)-\operatorname{adj} A \qquad (**)$$ for some polynomial $b(\lambda).$ Specializing $\lambda$ to $A,$ we conclude that

$$q(A)=-\operatorname{adj} A\qquad \square$$