Amazingly, this is exactly Lemma 3 in my paper "Almost all integer matrices have no integer eigenvalues" with Erick B. Wong (other than restricting to $i=1,j=2$ which is trivially generalized). There's a short proof there which seems to fit your needs. There's also a reference to a book on curious determinant identities of this sort in the bibliography (which is where we got the proof in the first place).

This proof is direct, which is based on the fact that the determinant of a square matrix can be computed by expansion about *any* column.

I will prove that $P(n)$ (in the description of the question) is true for $n = 2$, $3$, $\dots$ by mathematical induction.

It is easy to check that $P(2)$ is true.

Suppose that $P(n-1)$ is true (in which $n \geq 3$). I will prove that $P(n)$ is true under the hypothesis.

Choose a positive integer $q$ less than or equal to $n$. Choose a positive integer $q$ less than $p$. Suppose that column $p$ of the $n \times n$ matrix $A$ is equal to column $q$ of $A$.

**Choose a positive integer $u$ less than or equal to $n$ so that $u \neq p$ and that $u \neq q$.** Note that $A(i|u)$ has two equal columns (for $i = 1$, $2$, $\dots$, $n$). By the hypothesis, $\det {(A(i|u))} = 0$. Hence
$$
\begin{aligned}
\det {(A)}
= {} &
\sum_{i = 1}^{n} {(-1)^{i+u} [A]_{i,u} \det {(A(i|u))}}
\\
= {} &
\sum_{i = 1}^{n} {(-1)^{i+u} [A]_{i,u} \,0}
\\
= {} & 0.
\end{aligned}
$$

Hence by mathematical induction, $P(n)$ is true for $n = 2$, $3$, $\dots$.

## Best Answer

Note that the eigenvalues of $I + vv^T$ are all ones except for one of them which is $1 + v^T v = 1 + \Vert v \Vert^2$. Combine this with the definition of determinant being the product of all eigenvalues.