Question
Find the characteristic polynomial of
$$A = \begin{pmatrix} 1 & 2 & \cdots & n \\ n+1 & n+2 & \cdots & 2n \\ \vdots & \vdots & \ddots & \vdots \\ n^2-n+1 & n^2-n+2 & \cdots & n^2 \end{pmatrix}.$$
My approach
Observe that $A$ has rank $2$, and $$ A (1,1,\ldots,1)^t = (\frac{n(n+1)}{2}-n^2)(1,1,\ldots,1)^t + n^2(1,2,\ldots,n)^t $$ $$ A (1,2,\ldots,n)^t = (\frac{n(n+1)(2n+1)}{6}-\frac{n^2(n+1)}{2})(1,1,\ldots,1)^t + \frac{n^2(n+1)}{2}(1,2,\ldots,n)^t $$ So $\text{span}\{ (1,1,\ldots,1), (1,2,\ldots,n) \}$ is $L_A$-invariant. Then \begin{align*} \text{det}(A-tI) &= (-1)^n t^{n-2} \text{det} \begin{pmatrix} \frac{n(n+1)}{2}-n^2 - t & \frac{n(n+1)(2n+1)}{6}-\frac{n^2(n+1)}{2} \\ n^2 & \frac{n^2(n+1)}{2}-t \end{pmatrix} \\ &= (-1)^n t^{n-2} \left(t^2 - \frac{n+n^3}{2}t + \frac{n^3-n^5}{12} \right) \end{align*}
However the calculation get complicated. Is there any simpler way to do this?