Characteristic polynomial of a matrix is monic? Given a $n \times n$ matrix A, I need to show that its characteristic polynomial, defined as $P_A (x) = det (xI-A)$ is monic. I am trying induction. But no clue after induction hypothesis.

For $n=1$,
\begin{equation}
P_A(x)=det(xI_1-A)=|s-a_{11}|=s-a_{11}.
\end{equation}
So, $P_A(x)$ is monic for $n=1$.
Assume $P_A(x)$ is monic for $n=k$, i.e.
\begin{equation}
P_A(x)=det(xI_k-A)=\begin{vmatrix}
                   (x-a_{11})&-a_{12}&...&-a_{1k}\\
                   -a_{21}&(x-a_{22})&...&-a_{2k}\\
                   \vdots&\vdots&\ddots&\vdots\\
                   -a_{k1}&-a_{k2}&...&(x-a_{kk})
                   \end{vmatrix}=x^k+...
\end{equation}
We are now left to show that; $P_A(x)$ is monic for $n=k+1$, so
\begin{equation}
\begin{split}
P_A(x)=&det(xI_{k+1}-A)\\
=&\begin{vmatrix}
                   (x-a_{11})&-a_{12}&...&-a_{1k}&-a_{1,k+1}\\
                   -a_{21}&(x-a_{22})&...&-a_{2k}&-a_{2,k+1}\\
                   \vdots&\vdots&\ddots&\vdots\\
                   -a_{k1}&-a_{k2}&...&(x-a_{kk})&-a_{1,k+1}\\
                   -a_{k+1,1}&-a_{k+1,2}&...&-a_{k+1,k}&(x-a_{k+1,k+1})
\end{vmatrix}\\
=&\underbrace{(-a_{k+1,1})(-1)^{(k+1)+1}\begin{vmatrix}
                   -a_{12}&...&-a_{1k}&-a_{1,k+1}\\
                   (x-a_{22})&...&-a_{2k}&-a_{2,k+1}\\
                   \vdots&\ddots&\vdots\\
                   -a_{k2}&...&(x-a_{kk})&-a_{1,k+1}
                   \end{vmatrix}}_{order\ of\ x\ is\ k-1}\\
+&\underbrace{(-a_{k+1,2})(-1)^{(k+1)+2}\begin{vmatrix}
                   (x-a_{11})&-a_{13}&...&-a_{1k}&-a_{1,k+1}\\
                   -a_{21}&-a_{23}&...&-a_{2k}&-a_{2,k+1}\\
                   -a_{31}&(x-a_{33})&...&-a_{3k}&-a_{3,k+1}\\
                   \vdots&\ddots&\vdots\\
                   -a_{k1}&-a_{k3}&...&(x-a_{kk})&-a_{1,k+1}
                   \end{vmatrix}}_{order\ of\ x\ is\ k-1}\\
+&...\\
+&(x-a_{k+1,k+1})(-1)^{\overbrace{(k+1)+(k+1)}^{2(k+1)}}det(xI_k-A)\\
=&(x-a_{k+1,k+1})(x^{k}+...)+...\\
=&x^{k+1}+...
\end{split}
\end{equation}
So, $P_A(x)$ is also found monic for $n=k+1$.
$\therefore$ $P_A(x)$ is monic.$\square$
