Find the determinant of $I + A$

Here is a proof using eigenvalues, wishing you have already seen this notion.

Clearly, as all the columns of $A$ are proportional, the kernel has dimension $n-1$, thus eigenvalue $0$ is of order $n-1$.

The sum of the eigenvalues being equal to the trace $\sum_i a_i$, the $n$-th eigenvalue is:

$$\lambda:=\sum_i a_i.$$

Thus as the spectrum of $A+xI$ is equal to the spectrum of $A$ shifted by x (here with $x=1$), it is constituted of $1$ (with multiplicity $n-1$) and $\lambda+1$.

The determinant of a matrix being equal to the product of its eigenvalues, the final result is:

$$\det(A+I)=1+\sum_i a_i$$

Using Sylvester's determinant identity,

$$\det \left(\mathrm I_n + 1_n \mathrm a^{\top} \right) = \det \left( 1 + \mathrm a^{\top} 1_n \right) = 1 + \mathrm a^{\top} 1_n = 1 + \sum_{i=1}^n a_i$$

First look at what happens when $n=2$, $n=3$, renaming our matrix $B_n(a_1,\cdots,a_n)$ one gets

$$\begin{align}\det{B_2(a_1,a_2)}&=1+a_1+a_2\\\det{B_3(a_1,a_2,a_3)}&=1+a_1+a_2+a_3\end{align}$$

Assume that $\det{B_{n-1}(a_1,\cdots,a_{n-1})}=1+a_1+\cdots a_{n-1}$ and consider

$$\begin{vmatrix}1+a_1 & a_2 & \cdots & a_n\\ a_1 & 1+a_2 & \cdots & a_n\\ \vdots & \vdots & \ddots & \vdots\\ a_1 & a_2 & \cdots & 1+a_n \end{vmatrix}$$

Substract the second row from the first to get

$$\begin{vmatrix} 1 & -1 & 0 & \cdots & 0\\ a_1 & 1+a_2 & a_3 &\cdots & a_n\\ a_1 & a_2 & 1+a_3 & \cdots & a_n\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_1 & a_2 & a_3 &\cdots & 1+a_n \end{vmatrix}$$

Then add the first column to the second and get

$$\begin{vmatrix} 1 & 0 & 0 & \cdots & 0\\ a_1 & 1+a_1+a_2 & a_3 &\cdots & a_n\\ a_1 & a_1+a_2 & 1+a_3 & \cdots & a_n\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_1 & a_1+a_2 & a_3 &\cdots & 1+a_n \end{vmatrix}$$

Developing along the first row one gets

$$\det{B_n(a_1,\cdots,a_n)}=\det{B_{n-1}(a_1+a_2,\cdots ,a_n)}$$

And this by the induction assumption is $$1+a_1+a_2+\cdots+a_n$$