Determinant calculation Prove:
$$ \det\left[
\begin{array}{cccc}
1+x_1y_1 & x_1y_2 & \cdots & x_1y_n\\
 x_2y_1 & 1+x_2y_2  &  \cdots & x_2y_n \\
\vdots & \vdots & \ddots & \vdots  \\ 
 x_ny_1 & x_ny_2  &  \cdots & 1+x_ny_n \\
\end{array}
\right]=1+\sum_{i=1}^{n}x_iy_i$$
I tried to do some elementary operations, and develop by first row, but couldn't get further. 
Need to be proven without using eigenvalues.
Any help appreciated.
 A: The determinant can be written under the form:
$\det(I_n+UV^T)$ where $U=\left(\begin{array}{c}x_1\\x_2\\ \vdots \\x_n\end{array}\right)$ and $V=\left(\begin{array}{c}y_1\\y_2\\ \vdots \\y_n\end{array}\right).$
Let us recall that the matrix determinant lemma (https://en.wikipedia.org/wiki/Matrix_determinant_lemma) says that
$$\det(A+UV^T)=(1+V^TA^{-1}U) \det(A)$$
Taking $A=I$ gives:
$$\det(I_n+UV^T)=1+V^TU$$
which is the desired result.
A: Use $n$-linearity with respect to the first column to  put the first $1$ alone on the first column.
A: The matrix $A=(x_1,\dots,x_n)^T(y_1,\dots,y_n)$ is rank 1, so its characteristic polynomial is $X^n-\text{tr}(A)X^{n-1}$. Put $X=-1.$
Question has changed since this answer!
A: If you just take a deep breath and calculate it will come out! 
For after all, the answer is clearly going to be a sum of terms each of which is a product of $k$  $x$'s and $k$ $y$'s, for $k=0,1,\dots,n$. Every such term will arise as a product of $n-k$ $1$'s from the diagonal, multiplied by the top degree term of its cofactor, that is, multiplied by $\det\hat{x}^T\hat{y}$, where $\hat{x}$ is just the $k$-vector got from some $k$ of the indices. But now this little matrix is rank $1$ and so singular for all $k\not=0,1$. The only non-zero terms are then
$1$ and $\sum x_iy_i$. 
