Full-rank of an (almost) diagnoal matrix Let
$$
A =\begin{pmatrix}
a_1  & 1 & {\bf 0} &  {\bf 0}  \\ 
a_2 & \ddots & \ddots & {\bf 0}\\
\vdots & \ddots & a_1 & 1 \\
a_{n} & \ldots & a_2 & a_1
\end{pmatrix}
$$
I now want to know under what condition is $\text{Rank}(A)=n$.
The way I will know that $\text{Rank}(A)=n$ is using the fact that simple row and column operations don't change the rank. First, eliminate the element in the first row and first column using the $1$ in the top row, then eliminate the element in the second row and first column using the $1$ in the second row and so on up to and including eliminating the element in the $(n-1)^{th}$ row and first column. Now, for $\text{Rank}(A)=n$ the element in the first column and $n^{th}$ row cannot be zero. I get (if I haven't made a mistake in my algebra) the condition 
$$
a_n-a_{n-1}a_1-a_{n-2}(a_2-a_1^2)-a_{n-3}(a_3-2a_1a_2+a_1^3) - \ldots \neq 0
$$ 
Is there a simple way to explicitly express this inequality i.e. replace the "$\ldots$" above with something explicit? 

Not sure if there's any simpler solution method, but there is a closed-form formula for the matrix's determinant. See Alfred Inselberg (1978), On determinants of Toeplitz-Hessenberg matrices arising in power series, Journal of Mathematical Analysis and Applications, 63(2):347–353.
