Let A be an $n*n$ matrix. Prove that if $rank(A) = 1$, then $det(A + E) = 1 + trace(A)$ I feel like I've got the answer, but I've never been good at putting what I think into words.
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = 0 = n_{11}n_{22} - n_{12}n_{21}$
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}n_{22} + n_{11} + n_{22} + 1 - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{11} + n_{22} + 1 = n_{11} + n_{22} + 1$
Which shows that it's true for a 2*2 matrix.
Looking at other 2*2 matrices
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = n_{11}n_{22} + n_{22} - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{22} = n_{22}$
Similarly:
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}$
$\begin{vmatrix}
n_{11} & n_{12} + 1 \\
n_{21} & n_{22}
\end{vmatrix} = -n_{21}$
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} + 1 & n_{22}
\end{vmatrix} = -n_{12}$
Positive if on the main diagonal, negative if on the side diagonal.
Using that, we can show that
$\begin{vmatrix}
n_{11} + 1 & n_{12} & n_{13} \\
n_{21} & n_{22} + 1 & n_{23} \\
n_{31} & n_{32} & n_{33} + 1
\end{vmatrix} = (n_{11} + 1)(n_{22} + n_{33} + 1) - n_{12}n_{21} - n_{13}n_{31} = n_{11}n_{22} - n_{12}n_{21} + n_{11}n_{33} - n_{13}n_{31} + n_{11} + n_{22} + n_{33} + 1 = n_{11} + n_{22} + n_{33} + 1$
And in a similar fashion we can apply this to higher order matrices.
But I haven't the slightest clue how to word or show this "similar fashion".

Suppose $\;A=(a_{ij})\;$ with rank$\,A=1\;$ and assume that the only row in $\;A\;$ which is linearly independent (= non all-zeros row) is the first one, so that the other ones are scalar multiples of this first one. Say the $\;i\,-$ th row is $\;k_i\;$ times the first one
$$|A+E|=\begin{vmatrix}1+a_{11}&a_{12}&\ldots&a_{1n}\\a_{21}&1+a_{22}&\ldots&a_{2n}\\\ldots&\ldots&\ldots&\ldots\\a_{n1}&a_{n_2}&\ldots&1+a_{nn}\end{vmatrix}=$$
We now apply elementary operations on the above which do not change the determinant: from row $\;i\;$ substract $\;k_i\;$ times the first row:
$$=\begin{vmatrix}1+a_{11}&a_{12}&a_{13}&\ldots&a_{1n}\\-k_2&1&0&\ldots&0\\-k_3&0&1&0\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\-k_n&0&0&\ldots&1\end{vmatrix}\stackrel{\text{develop 2nd row}}=k_2\begin{vmatrix}a_{12}&a_{13}&\ldots&a_{1(n-1)}&a_{1n}\\0&1&0&\ldots&0\\0&0&1&\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\0&0&0&\ldots&1\end{vmatrix}+$$$${}$$
$$\begin{vmatrix}1+a_{11}&a_{13}&\ldots&a_{1(n-1)}&a_{1n}\\-k_3&1&0&\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\-k_n&0&0&\ldots&1\end{vmatrix}\stackrel{\text{Inductively}}=$$$${}$$
$$=k_2a_{12}+k_3a_{13}+\ldots+k_{n-1}a_{1(n-1)}+k_na_{1n}+1+a_{11}=$$
$$=1+a_{11}+a_{22}+\ldots+a_{nn}=1+\text{Tr.}\,A$$
Try to fill in details. After some thought, if you still have doubts write back.
