You matrix $A = (a_{ij})$ is
an upper triangular matrix
( $a_{ij} = 0$ whenever $i > j$ )
and a Toeplitz matrix
( $a_{ij}$ depends only on $i-j$ ) at the same time.
I cannot find any reference online which teach you how to evaluate its inverse effectively.
I hope this keywords can help you in your own search.
If you just want an inverse without too many other concerns, it is actually
pretty easy to get the inverse ourselves.
Let $\eta$ be the $n \times n$ matrix with $1$ on its superdiagonal and $0$ otherwise. i.e.
$$\eta = (\eta_{ij}),\quad \eta_{ij} = \begin{cases}1,& i - j = -1\\0,& \text{otherwise}\end{cases}$$
We have $\eta^n = 0$ and we can express $A$ as a polynomial in $\eta$.
$$A = x_1 I + x_2 \eta + x_3 \eta^2 + \cdots + x_n \eta^{n-1}$$
$A$ will be invertible when and only when $x_1$ is non-zero. When $A$ is invertible,
$A^{-1}$ is also an upper triangular Toeplitz matrix. We can also represent it as a polynomial in $\eta$.
Introduce numbers $\displaystyle\;\alpha_i = \frac{x_{i+1}}{x_1}$ and $\beta_i$ ( $i = 1,\ldots,n-1$ ) such that
$$\begin{align}
A &= x_1 \left(I + \alpha_1 \eta + \alpha_2 \eta^2 + \cdots + \alpha_{n-1} \eta^{n-1}\right)\\
A^{-1} &= x_1^{-1} \left(I + \beta_1 \eta + \beta_2 \eta^2 + \cdots + \beta_{n-1} \eta^{n-1}\right)
\end{align}
$$
The condition $A^{-1} A = I$ can be expanded to following set of relations. They will alow you to compute $\beta_k$ in a recursive manner.
$$\begin{align}
-\beta_1 &= \alpha_1\\
-\beta_2 &= \alpha_1 \beta_1 + \alpha_2\\
&\;\vdots\\
-\beta_k &= \alpha_1 \beta_{k-1} + \alpha_2 \beta_{k-2} + \cdots + \alpha_k\\
&\;\vdots
\end{align}
$$
When $n$ is small and you want individual $\beta_k$ as a function of $\alpha_k$.
There is actually a trick to get it. You can ask a CAS to compute the Taylor
expansion of the reciprocal of following polynomial in $t$:
$$\frac{1}{1 + \alpha_1 t + \alpha_2 t^2 + \cdots + \alpha_{n-1} t^{n-1}}
= 1 + \beta_1 t + \beta_2 t^2 + \cdots + \beta_{n-1} t^{n-1} + O(t^n)$$
The coefficients of $t^k$ ($1 \le k < n$) in the resulting Taylor expansion will
be the expression you want for $\beta_k$. e.g.
$$\begin{align}
\beta_1 &= -\alpha_1,\\
\beta_2 &= \alpha_1^2 - \alpha_2,\\
\beta_3 &= -\alpha_1^3 + 2\alpha_1\alpha_2 - \alpha_3,\\
\beta_4 &= \alpha_1^4 - 3\alpha_1^2\alpha_2 + \alpha_2^2 + 2\alpha_1 \alpha_3 - \alpha_4\\
&\;\vdots
\end{align}$$
I assume that $\star$ is allowed to be zero.
We attain the minimal possible rank by setting each $\star = 0$. Any matrix in this pattern will necessarily have rank at least $2$ because we always have the rank $2$ submatrix
$$
\pmatrix{100&\star \\0 & 203}
$$
We attain the maximal possible rank by setting each $\star = 1$. Since the matrix is in row-echelon form, the rank is simply the number of leading non-zero entries, which is $n-1$. We cannot attain rank $n$ because the first column is always $0$.
It is possible to attain any rank in between by setting columns equal to $0$.
Best Answer
Strictly upper triangular matrices are nilpotent. Indeed, the characteristic polynomial function of such a matrix $S$ is given by $p(\lambda)=\lambda^n$ (since all the diagonal entries of the matrix are zero) whence by cayley-hamilton $S^n=0$. Hence $$ (I+S+S^2+\dotsb+S^{n-1})(I-S)=I-S^n=I $$ as desired.