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?

Here's a geometric proof: whether it's simpler/more intuitive than the row manipulations proof is a matter of taste.
Let $e_i$ be the elementary vector with a $1$ in the $i$th entry and $0$ everywhere else, and let $u$ denote the all-ones vector. We seek to find the determinant of the matrix $M$ whose $i$th row is $v_i:=u+e_i$. Since $M$ scales volumes by $\det M$, this determinant is the ratio
$$\det M=\frac{\operatorname{Volume}(\text{simplex with vertices }\{0,v_1,\dots,v_n\})}{\operatorname{Volume}(\text{simplex with vertices }\{0,e_1,\dots,e_n\})},$$
since $Me_i=v_i$. For simplex volumes, we use the formula $\frac1n\text{base}\cdot\text{height}$, with bases of each determined by $\{e_1,\dots,e_n\}$ and $\{v_1,\dots,v_n\}$. Since the $(n-1)$-simplex with vertices $\{v_1,\dots,v_n\}$ is just a translation of the simplex with vertices $\{e_1,\dots,e_n\}$, their bases have the same volume, and so $\det M$ is the ratio of the distances from $0$ to the planes determined by $\{v_1,\dots,v_n\}$ and $\{e_1,\dots,e_n\}$.
These planes are preserved under any permutation of the axes, and so the projection of $0$ down to each of these planes must be preserved by such transformations as well, and must be some scalar multiple of the all-ones vector $u$. The multiple of $u$ on the plane spanned by $\{e_1,\dots,e_n\}$ is $\frac1nu$, while the multiple of $u$ on the plane spanned by $\{v_1,\dots,v_n\}$ is $(1+\frac 1n)u$. So,
$$\det M=\frac{\left(1+\frac 1n\right)\|u\|}{\frac 1n\|u\|}=n+1.$$
