Matrix proof by induction Problem
Let $A_n\in M\left(\mathbb{R}\right)$ such that
$$A_n=\begin{pmatrix}1+x_1&x_2&...&x_n\\ x_1&1+x_2&...&x_n\\ ...&...&...&...\\ x_1&x_2&...&1+x_n\end{pmatrix}$$
Demonstrate that $\det(A_n)=1+x_1+x_2+...+x_n$
Thoughts
I think that the best way is by induction.
Starting with a inductive base
let say $n=2$
$$A_2=\begin{pmatrix}1+x_1&x_2\\ x_1&1+x_2\end{pmatrix}$$
$$\det(A_2)=1+x_2+x_1$$
Its true for that case
Now the inductive hypothesis would be for $n=k$
$$\det(A_k)=\begin{pmatrix}1+x_1&x_2&...&x_k\\ x_1&1+x_2&...&x_k\\ ...&...&...&...\\ x&x_2&...&1+x_k\end{pmatrix}=1+x_1+x_2+...+x_k$$
Now we need to prove that $n=k+1$
$$\det(A_n)=\begin{pmatrix}1+x_1&x_2&...&x_k&x_{k+1}\\ x_1&1+x_2&...&x_k&x_{k+1}\\ x_1&x_2&...&...&....\\ ...&...&...&1+x_k&...\\ x_1&x_2&...&x_k&1+_{k+1}\end{pmatrix}=1+x_1+x_2+...+x_{k+1}$$
Then of that I'm unsure about how to proceed, I tried something like representing them by their cofactors but didn't let me anywhere. Any help would be appreciated.
 A: If you subtract the second line of $A_n$ from the first one, the determinant remains the same and the entries of the first line become are $1,-1,0,0,\ldots,0$. Now, if you subtract the third line from the second one, the entries of the second line become $0,1,-1,0,\ldots,0$. And so on. So$$\det A_n=\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\x_1&x_2&x_3&x_4&\ldots&x_{n-1}&1+x_n\end{vmatrix}.\tag1$$Now, if you add the first line times $-x_1$ to the last line, the $RHS$ of $(1)$ becomes$$\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\0&x_1+x_2&x_3&x_4&\ldots&x_{n-1}&1+x_n\end{vmatrix}.\tag2$$And now, if you add the second lines times $-x_1-x_2$ to the last line, the RHS of $(2)$ becomes$$\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\0&0&x_1+x_2+x_3&x_4&\ldots&x_{n-1}&1+x_n\end{vmatrix},$$and so on… In the end, you will get$$\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\0&0&0&0&\ldots&0&1+x_1+\cdots+x_n\end{vmatrix}=1+x_1+\cdots+x_n.$$
