Help me to prove the determinant of given matrix. Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
      -x & a_2&a_3&a_4&\cdots &a_n\\
      a_{1}+x & -x-a_2 & 0&0&\cdots &0\\
      a_1+x&0 & -x-a_3 &0&\cdots &0\\
      \vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
      a_1+x&0 & 0&0&\cdots & -x-a_n\\
      \end{array}\end{bmatrix}$, then how to find the $\det (M)$? 
The empirical formula I got from considering $n=2,3,4$ in Wolfram Mathematica is
$$
(-1)^n(x^n-\sum_{k=2}^n (k-1)\sigma_k x^{n-k})
$$
where $\sigma_k$ is the $k$-th elementary symmetric polynomial in $a_1,\dots,a_n$.

I start with the well-known and easily proved fact that the $n\times n$ all $1$'s matrix 
$$
\begin{pmatrix}
1 & 1 & \dots &1\\
1& 1  & \dots &1\\
\vdots &\vdots &\ddots&\vdots\\
1 & 1 &\dots  &1
\end{pmatrix}
$$
has one eigenvalue $n$ corresponding to the obvious eigenvector  $(1,1,\dots,1)^{T}$ and the eigenvalue $0$ occurring $n-1$ times corresponding to the obvious eigenvectors $(1,-1,0,\dots, 0)^{T}$, $(1,0,-1,\dots, 0)^{T}$, ... $(1,0,0,\dots, -1)^{T}$. We will then also know the eigenvlaues of the matrix got by subtracting the identity matrix from the all-$1$'s matrix, and so we  will have that
$$
\det
\begin{pmatrix}
0 & 1 & \dots &1\\
1& 0  & \dots &1\\
\vdots &\vdots &\ddots&\vdots\\
1 & 1 &\dots  &0
\end{pmatrix}
=
(n-1)(-1)^{n-1}. 
$$
Now turn to the matrix $M$ of the question. To evaluate the determinant we can replace it by the matrix we get by adding the first row to every other row, that is by
$$
\begin{pmatrix}
-x & a_2 &\dots &a_n\\
a_1& -x & \dots & a_n\\
\vdots & \vdots &\ddots &\vdots\\
a_1 & a_2 &\dots & -x
\end{pmatrix}.
$$
The coefficient of $(-x)^{n-k}$ in this expansion is clearly the sum of all the principal $k\times k$ minors of 
$$
\begin{pmatrix}
0 & a_2 &\dots &a_n\\
a_1& 0 & \dots & a_n\\
\vdots & \vdots &\ddots &\vdots\\
a_1 & a_2 &\dots & 0
\end{pmatrix}.
$$
For $k=0$ that sum is $0$; otherwise it is the symmetric sum of terms like
$$
\det
\begin{pmatrix}
0 & a_2 & \dots &a_k\\
a_1& 0  & \dots &a_k\\
\vdots &\vdots &\ddots&\vdots\\
a_1& a_2 &\dots  &0
\end{pmatrix}
=
a_1 a_2\dots a_k \det
\begin{pmatrix}
0 & 1 & \dots &1\\
1& 0  & \dots &1\\
\vdots &\vdots &\ddots&\vdots\\
1& 1 &\dots  &0
\end{pmatrix}
=
a_1 a_2\dots a_k (-1)^{k} (k-1).
$$
These calculations will provide a proof of the OP's conjecture.
