Sorry, I'm having trouble proving something by using the Laplace expansion.
Let $K$ be a field and $a,b \in K^n$. Prove or disprove the following statement: $\det(E_n+ab^T)= 1+a^Tb$.
Let $a= \begin{bmatrix} a_1\\\vdots\\a_n \end{bmatrix}$ and $b=\begin{bmatrix} b_1\\\vdots\\b_n \end{bmatrix}$. I know that $a^Tb=(a_1b_1+a_2b_2+\dots a_nb_n)$ and $ab^T= \begin{bmatrix} a_1b_1 &a_1b_2 & \dots & a_1b_n \\a_2b_1&a_2b_2 &\dots &a_2b_n \\ \vdots & \vdots \\a_nb_1 &a_nb_2& \dots & a_nb_n\end{bmatrix}$.
We then have $E_n+ab^T= \begin{bmatrix} 1+a_1b_1 &a_1b_2 & \dots & a_1b_n \\a_2b_1& 1+a_2b_2 &\dots &a_2b_n \\ \vdots & \vdots \\a_nb_1 &a_nb_2& \dots & 1+a_nb_n\end{bmatrix}$ and also $1+a^Tb= 1+a_1b_1+a_2b_2+...a_nb_n$.
My intuition tells me that the statement is true but I haven't been able come up with full proof. After establishing these two equalities I thought about an attempt at proof by induction on $n$.
Base case: $\det(E_2+ab^T)=(1+a_1b_1)(1+a_2b_2)-(a_1b_2)(a_2b_1)=1+a_2b_2+a_1b_1+a_1b_1a_2b_2-a_1b_1a_2b_2=1+a^Tb.$ We then assume the statement holds true until $n$ and observe the $n+1$th case.
$\det(E_{n+1}+ab^T)=1+a^Tb$. Let $E_{n+1}+ab^T=M$ we then use the Laplace expansion: $$\det(M)= \sum_{j=1}^{n+1}(-1)^{i+j}M_{i,j}S_{i,j}$$ Where $S_{i,j}$ is determinant of the the submatrix obtained by removing the $i$th row and $j$th column. Therefore we have that $S_{i,j}= \det(E_n+ab^T)=1+a^Tb$ by our assumption so in the end:$$ \det(M)= \sum_{j=1}^{n+1}(-1)^{i+j}M_{i,j}(1+a^Tb)$$ This is currently where I'm stuck and to be honest I don't even know if what I've done is correct in the first place. I would aprreciate help.
