How to find the determinant of a square matrix of order n? I have the following matrix $M_n = \begin{pmatrix} 1+x & -1 &&& \\ 1&x&-1&& \\ \vdots &&\ddots&\ddots& \\1&&&\ddots&-1 \\1&&&&x\end{pmatrix}$.
So, I decided to use the multilinearity of the determinant then, $M_n =  \begin{pmatrix} 1 & -1 &&& \\ 1&x&-1&& \\ \vdots &&\ddots&\ddots& \\1&&&\ddots&-1 \\1&&&&x\end{pmatrix} + \begin{pmatrix} x & -1 &&& \\ 0&x&-1&& \\ \vdots &&\ddots&\ddots& \\0&&&\ddots&-1 \\0&&&&x\end{pmatrix} = A_n + B_n$
To simplify the problem, I think it's better to use the transpose of $A_n$ would be more practical, which give us : 
$$\det(M_n) = \begin{vmatrix} 1 &1&1&\ldots&1 \\ -1 &x&&& \\ &-1&\ddots& \\ &&\ddots&\ddots&\\ &&&-1&x\end{vmatrix} + \begin{vmatrix} x & -1 &&& \\ 0&x&-1&& \\ \vdots &&\ddots&\ddots& \\0&&&\ddots&-1 \\0&&&&x\end{vmatrix}\\ =\  ...\ = A_{n-1} + B_{n-1} + B_n = \sum_{k = 1}^n B_k +A_1$$ (by induction) $=\sum_{k = 0}^n x^k$
It is the first time, I doing such exercise. I would like to know if what I wrote would be accepted as a good one or should I use another method?
Thanks!
Edit : I forgot to say "Hello"
 A: Beware : your proof isn't valid because you use implicitly a formula 
$$\det(A+B)=\det(A)+\det(B)$$ 
which is erroneous. It's by chance that you get the right results.
Make instead a Laplace expansion of $D_n:=\det(M_n)$ with respect to its last column: 
$$D_n=x\color{red}{(+1)}D_{n-1}+(-1)\color{red}{(-1)}\Delta \tag{1}$$
(the signs changing a minor into a cofactor are highlighted in $\color{red}{red}$) where $\Delta$ is a $(n-1) \times (n-1)$ determinant with last row  $(1,0,\cdots 0)$. Expanding this determinant with respect to this row, you will get $\color{red}{(-1)^{n-2}}$ times a $(n-2) \times (n-2)$ matrix which is lower triangular, therefore whose determinant is the product of its diagonal elements, i.e., $(-1)^{n-2}$.
Consequently, (1) becomes :
$$D_n=xD_{n-1}+(-1)^{n-2}(-1)^{n-2} \ \ \iff \ \ D_n=xD_{n-1}+1\tag{2}$$
(with $D_1=x+1$). By a straightforward induction :
$$D_n=x(x\cdots (x(x+1)+1)\cdots)+1)+1\tag{3}$$
$$D_n=x^n+x^{n-1}+x^{n-2}+\cdots +x+1 \tag{4}$$
Remarks : 
1) if you happen to know the concept of companion matrix, there is an easy alternative proof using the fact that $det(-M_n)$ is the characteristic polynomial of the companion matrix of the polynomial in the RHS of (4). 
2) (3) can be called Horner factorization of (4).
