Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
  1 & 0 & 0 &\cdots & 0 \\
  0 & 1 & 0 &\cdots & 0 \\
  0 & 0 & 1 &\cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & 0 & \cdots & 1
 \end{pmatrix} +\begin{pmatrix}
  1 & 1 & 1 & \cdots & 1 \\
  1 & 1 & 1 & \cdots & 1 \\
  1 & 1 & 1 & \cdots & 1 \\
  \vdots &  \vdots & \vdots & \ddots & \vdots \\
  1 & 1 & 1 &  \cdots & 1
 \end{pmatrix}\right| =\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  1 & 2 & 1 & \cdots & 1 \\
  1 & 1 & 2 & \cdots & 1 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & 1 & 1 & \cdots & 2
 \end{matrix} \right|=n+1.$$
I can prove it by adding and subtracting rows and columns without changing the determinant:

*

*subtracting the first row from the others $
\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  1 & 2 & 1 & \cdots & 1 \\
  1 & 1 & 2 & \cdots & 1 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & 1 & 1 & \cdots & 2
 \end{matrix} \right|=
\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|$


*subtracting the other rows from the first $\left|\begin{matrix}
  2 & 1 & 1 & \cdots & 1 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|=
\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|$


*adding the other columns to the first $\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  -1 & 1 & 0 & \cdots & 0 \\
  -1 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  -1 & 0 & 0 & \cdots & 1
 \end{matrix} \right|=
\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  0 & 1 & 0 & \cdots & 0 \\
  0 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & 0 & \cdots & 1
 \end{matrix} \right|$


*and clearly $\left|\begin{matrix}
  n+1 & 0 & 0 & \cdots & 0 \\
  0 & 1 & 0 & \cdots & 0 \\
  0 & 0 & 1 & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & 0 & \cdots & 1
 \end{matrix} \right| =n+1$
but I feel there should be something more intuitive.
What would be a more intuitive or insightful way to prove this?

I'm going to elaborate on Exodd's comment. I'm inclined to agree that this is the simplest proof, provided you're comfortable with the basic constructions of linear algebra.
Lemma: If $A \in \mathbf{C}^{n \times n}$ has eigenvalues (counting multiplicity) $(\lambda_i)_{i=1}^n$, then $A + I$ has eigenvalues $(\lambda_i + 1)_{i=1}^n$.
Proof: Observe that $\det(xI - (A + I)) = \det((x-1)I - A)$. $\square$

Proposition: If $J \in \mathbf{C}^{n \times n}$ is the all-ones matrix, then $\det(I+J) = n+1$.
Proof: Since $J$ is rank 1, hence it has exactly one nonzero eigenvalue. Since $\text{trace}(J) = \sum_{i=1}^n \lambda_i$, it follows that the unique nonzero eigenvalue $\lambda_1 = n$. By the lemma, $\det(J+I) = \prod_{i=1}^n (\lambda_i + 1) = n+1$. $\square$
