Is it true that $\det(A + I) = \operatorname{trace} (A) + 1$? 
Let $A$ be an $n\times n$ matrix such that $\operatorname{rank} A = 1$ and that $n- 2$ rows of $A$
  are the zero rows. Is it true that $\det(A + I) = \operatorname{trace} (A) + 1$?

I already have the answer which can be viewed here.
I don't understand how they expanded the determinant and got: $\begin{vmatrix}
a_i+1 & a_k  \\
\beta a_i & \beta a_k+1
\end{vmatrix}$
.
I saw another answer where he divided the determinant to two upper diagonals matrices but I didn't get how he did that. 
An explaination of these two answers and/or another way to solve it would be appreciated.
NOTE: we haven't covered eigenvalues.
 A: Since $\operatorname{rank}(A) =1$, $\lambda=0$ is an eigenvalue of multiplicity $n-1$.
As the sum of the eigenvalues is $\operatorname{tr}(A)$, it follows that the remaining eigenvalue is $\operatorname{tr}(A)$.
Then the eigenvalues of $A+I$ are $\lambda=0+1$ with multiplicity $n-1$ and $\lambda=\operatorname{tr}(A)+1$ with multiplicity $1$. Thus, $\det(A+I)$ is their product, that is
$$\det(A+I)=(0+1)^{n-1} (\operatorname{tr}(A)+1) \,.$$
A: Here is a solution without eigenvalues, which I think is exactly the solution you posted with the missing details:
Since the matrix has rank 1, all rows are multiple of one of the rows. Without loss of generality let all rows be mulple of the first row (otherwise interchange the first and kth row AND columns).
The rows of $A$ are $r_1, \alpha_2 r_1,..., \alpha_n r_1$.
Let $r_1=\begin{bmatrix}a_1,a_2,a_3,...,a_n\end{bmatrix}$. Then
$$tr(A)=a_1+\alpha_2 a_2+...+ \alpha_n a_n \,.$$
Now, when calculating $\det(A+I)$ make the following row operations: $R_2-\alpha_2 R_1, R_3-\alpha_3 R_1,...,R_n-\alpha_n R_1$. Then you get
$$\det(A+I)= \det\begin{bmatrix}a_1+1&a_2&a_3&...&a_n\\
\alpha_2a_1&\alpha_2a_2+1&\alpha_2a_3&...&\alpha_2a_n\\
\alpha_3a_1&\alpha_3a_2&\alpha_3a_3+1&...&\alpha_3a_n\\
...&...&...&...&...\\
\alpha_na_1&\alpha_na_2&\alpha_na_3&...&\alpha_na_n+1\\
\end{bmatrix}\\
= \det\begin{bmatrix}a_1+1&a_2&a_3&...&a_n\\
-\alpha_2&1&0&...&0\\
-\alpha_3&0&1&...&0\\
...&...&...&...&...\\
-\alpha_n&0&0&...&1\\
\end{bmatrix}$$
Now, expanding the determinant by the first row we get:
$$\det(A+I) 
= (a_1+1)\det\begin{bmatrix}
1&0&...&0\\
0&1&...&0\\
...&...&...&...\\
0&0&...&1\\
\end{bmatrix}-a_2\det\begin{bmatrix}
-\alpha_2&0&...&0\\
-\alpha_3&1&...&0\\
...&...&...&...\\
-\alpha_n&&0&...&1\\
\end{bmatrix}\\+a_3\det\begin{bmatrix}
-\alpha_2&1&...&0\\
-\alpha_3&0&...&0\\
...&...&...&...\\
-\alpha_n&0&...&1\\
\end{bmatrix}+...+(-1)^na_n\det\begin{bmatrix}
-\alpha_2&1&0&...&0\\
-\alpha_3&0&1&...&0\\
...&...&...&...&...\\
-\alpha_{n-1}&0&0&...&1\\
-\alpha_n&0&0&...&0\\
\end{bmatrix}$$
All these determinants are easy to calculate, and you get
$$\det(A+I)=a_1+1+\alpha_2 a_2+...+ \alpha_n a_n =1+tr(A)$$
A: This is actually true for any rank $1$ matrix regardless of the zero rows condition. One way to see this is that $0$ is an eigenvalue of $A$ with geometric multiplicity $n-1$.  The remaining eigenvalue must be equal to $\text{tr} (A)$ (since the sum of the eigenvalues is the trace).  The eigenvalues of $A$ are $0, 0, \ldots, 0, \text{tr}(A)$, and the eigenvalues of $A+I$ are shifted up by $1$, so: $1,1,\ldots, 1, 1+\text{tr}(A)$.  The determinant is just the product of these, which is $1+\text{tr}(A)$.
Alternative solution without eigenvalues: You have probably seen by now that expanding the determinant along a row with only one non-zero entry is particularly easy: instead of a sum, you only get one term.  For example, expanding on the first row of the below matrix gives:
$$\left|\begin{matrix}\color{red}0&\color{red}3&\color{red}0 \\ a & b & c \\ d & e  & f\end{matrix}\right| = (-1)^{1+2} \cdot 3 \cdot \left|\begin{matrix} a & c \\ d & f\end{matrix}\right|,$$
where the exponent $1+2$ comes from the row/column of the non-zero element, and the $3$ is the value of that element.
Things get even easier when you have a row whose only non-zero element is a $1$ situated anywhere on the main diagonal (i.e. row $i$ and column $i$ for some $i$).  In this case, the term $(-1)^{i+j} = (-1)^{i+i} = (-1)^{2i}$ simplifies to $1$, and the value of the non-zero element is $1$, so the determinant of the larger matrix is exactly equal to the submatrix obtained by deleting row $i$ and column $i$, e.g.:
$$\left|\begin{matrix}a & b & c \\ \color{red}0&\color{red}1&\color{red}0 \\ d & e  & f\end{matrix}\right| =  \left|\begin{matrix} a & c \\ d & f\end{matrix}\right|.$$
Now, think very carefully about the structure of the matrix $A+I$ in the solution you linked to.  This contains a large number of rows with the same convenient property as the one we have above: every element is equal to zero except for a single $1$ on the main diagonal.  The determinant of this matrix reduces to the determinant of the smaller matrix obtained by deleting the row and column of any of these solitary $1$'s.  Now try to picture what happens as you systematically eliminate each one: in the end, only two rows and two columns remain: the rows $i$ and $k$ and the columns $i$ and $k$.
This is how they reach the small matrix $\left|\begin{matrix} \alpha_i+1 & \alpha_k \\ \beta \alpha_i & \beta \alpha_k + 1\end{matrix}\right|$, just by extracting those entries from the larger matrix.  If you don't understand where the larger matrix comes from, please state so in the question itself.
A: The equality can be viewed as a consequence of Sylvester's determinant theorem $\det(I+XY)=\det(I+YX)$ (whose proof is merely based on the identity $\det(PQ)=\det(P)\det(Q)$ for square matrices) and the cyclic-ness of trace (i.e. $\operatorname{trace}(XY)=\operatorname{trace}(YX)$; the proof of this equality is straightforward --- just compare both sides entrywise). Since $A$ has rank 1, $A=uv^T$ for some vectors $u$ and $v$. Therefore
$$
\det(A+I_n)=\det(uv^T+I_n)=\det(v^Tu+I_1)=v^Tu+1=\operatorname{trace}(v^Tu)+1=\operatorname{trace}(uv^T)+1=\operatorname{trace}(A)+1.
$$
