Characteristic polynomial of the matrix with zeros on the diagonal and ones elsewhere

Find the characteristic polynomial of the matrix with zeros on the diagonal and ones elsewhere.

I've been able (I believe) to guess how it looks like (by considering matrices of small orders): $(x-n+1)(x-1)^{n-1}$. I suppose I should prove it by induction. But I don't know how to obtain the characteristic polynomial of a matrix of order $n+1$ from that of a matrix of order $n$ (i.e., how to make the inductive step).

Other methods of solution are also welcome. (Is it possible to use row reduction?)

Let $M$ be the matrix in question. Then $M=J-I$ where $J$ is the all-ones matrix. Then $$\chi_M(t)=\det(tI-M)=\det(tI+I-J)=\chi_J(t+1).$$ We just need to compute the characteristic polynomial of $J$, But $J^2=nJ$, so the eigenvalues of $J$ are in the set $\{0,n\}$. The trace of $J$ is $n$, so that one eigenvalue of $J$ is $n$ and the other $n-1$ are all zero. That is, $\chi_J(t)=t^{n-1}(t-n)$. Then $$\chi_M(t)=(t+1)^{n-1}(t-n+1).$$

Let the matrix be $A$. It's much easier to find the characteristic polynomial of $B=A+I$, the matrix that is all $1$s: we can do this by finding the eigenvectors, which will give us the eigenvalues. Then the eigenvalues of $A$ will just be those of $B$ with a $-1$, since $Bx=\lambda x \iff Ax = (\lambda-1)x$.

So what are the eigenvectors of $B$? It's so symmetric that it's easy to write a few down: the vector that is all $1$s has eigenvalue $n$, while any vector whose components sum to zero is an eigenvector with eigenvalue $0$. There are at least $n-1$ of these, since $e_1-e_i$ for $2 \leq i \leq n$ is a linearly independent subset of the eigenspace. Thus $0$ is an eigenvalue with multiplicity at least $n-1$, but we know the other eigenvalue already, so this is everything. Hence the characteristic polynomial is $(t-n)t^{n-1}$.

Returning to $A$, we find that this gives the characteristic polynomial of $A$ as $(t-n+1)(t+1)^{n-1}$.