Let
\begin{equation\*}
\begin{split}
M\_m
&=\begin{pmatrix}
-\binom{1}{0} & \binom{2}{0} &-\binom{3}{0} &\dotsm & (-1)^{m-1}\binom{m-1}{0} & (-1)^m\binom{m}{0}\\
0 & \binom{2}{1} &-\binom{3}{1} &\dotsm & (-1)^{m-1}\binom{m-1}{1} & (-1)^m\binom{m}{1}\\
0 & 0 &-\binom{3}{2} &\dotsm & (-1)^{m-1}\binom{m-1}{2} & (-1)^m\binom{m}{2}\\
\vdots & \vdots &\vdots &\ddots & \vdots & \vdots\\
0 & 0 & 0 &\dotsm & (-1)^{m-1}\binom{m-1}{m-2} & (-1)^m\binom{m}{m-2}\\
0 & 0 & 0 &\dotsm & 0 & (-1)^m\binom{m}{m-1}
\end{pmatrix}\_{m\times m}\\
&=(M\_{i,j})\_{m\times m},
\end{split}
\end{equation\*}
where
\begin{equation\*}
M\_{i,j}=
\begin{cases}
(-1)^{j}\dbinom{j}{i-1}, & 1\le i\le j\le m;\\
0, & 1\le j<i\le m.
\end{cases}
\end{equation\*}
For $m=5$, by the famous software Mathematica, we obtain
$$
\begin{pmatrix}
 -1 & 1 & -1 & 1 & -1 \\
 0 & 2 & -3 & 4 & -5 \\
 0 & 0 & -3 & 6 & -10 \\
 0 & 0 & 0 & 4 & -10 \\
 0 & 0 & 0 & 0 & -5 \\
\end{pmatrix}^{-1}
=
\begin{pmatrix}
 -1 & \frac{1}{2} & -\frac{1}{6} & 0 & \frac{1}{30} \\
 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} & 0 \\
 0 & 0 & -\frac{1}{3} & \frac{1}{2} & -\frac{1}{3} \\
 0 & 0 & 0 & \frac{1}{4} & -\frac{1}{2} \\
 0 & 0 & 0 & 0 & -\frac{1}{5} \\
\end{pmatrix}.
$$
What is the inverse of the triangular matrix $M\_m$ for $m\in\mathbb{N}=\{1,2,\dotsc\}$?


The matrix $M\_m$ comes from the recursive relation
\begin{equation}\label{beta(m+1minus1)}
\sum\_{k=j+1}^{m}(-1)^{k}\binom{k}{j}\beta\_{m+1,k}
=(-1)^{j+1} \beta\_{m,j}, \quad 0\le j\le m-1,
\end{equation}
where the first few $\beta\_{m,j}$ are
\begin{align\*}
\beta\_{1,0}&=1, & & & & & &\\
\beta\_{2,0}&=\frac{5}{3}, & \beta\_{2,1}&=1, & & & &\\
\beta\_{3,0}&=\frac{11}{5}, & \beta\_{3,1}&=\frac{13}{6}, & \beta\_{3,2}&=\frac{1}{2}, & &\\
\beta\_{4,0}&=\frac{93}{35}, & \beta\_{4,1}&=\frac{101}{30}, & \beta\_{4,2}&=\frac{4}{3}, & \beta\_{4,3}&=\frac{1}{6}.
\end{align\*}
We can also derive
\begin{align\*}%\label{beta(m+1)m-form}
\beta\_{m,m-1}&=\frac{1}{(m-1)!}, \quad m\ge1,\\
\beta\_{m,m-2}&=\frac{3m+4}{6(m-2)!}, \quad m\ge2,\\
\beta\_{m,m-3}&=\frac{15 m^2+35 m+24}{120(m-3)!}, \quad m\ge3,\\
\beta\_{m,m-4}&=\frac{105 m^3+315 m^2+364 m+176}{5040(m-4)!}, \quad m\ge4.
\end{align\*}
We guess that
\begin{equation\*}
\beta\_{m,m-k}=\frac{1}{(2k-1)!(m-k)!}\sum\_{\ell=0}^{k-1}\theta\_{k,\ell} m^\ell, \quad m\ge k,
\end{equation\*}
where $\theta\_{k,\ell}$ is a sequence of positive integers.


What is the explicit or closed-form expression of the sequence $\theta\_{k,\ell}$ for $0\le\ell\le n-1$?
What is the explicit or closed-form expression of the sequence $\beta\_{m,j}$ for $0\le j\le m-1$?



What is the inverse of a triangular matrix whose nonzero elements are binomial coefficients? What is the closed-form solution to a recursive relation?
This may answer the question, the sequence is the only implicit thing. I consider $Q=\begin{pmatrix}Q\_{i,j}\end{pmatrix}\_{n\times n}$ for $n\ge 3$, where
$$Q\_{i,j}=
\begin{cases}
\dbinom{j}{i-1}, & 1\le i\le j\le n;\\
0, & 1\le j<i\le n.
\end{cases}
$$
This is the same as the one defined in the question up to multiplying it by a $\pm 1$ diagonal matrix.
Define sequence $u\_k$ by $u\_1=1$ and
$$
u\_k=-\sum\_{i=0}^{k-2}\dfrac{u\_{i+1}}{(k-i)!}
$$
for every $k>1$. The $n\times n $ matrix $Q^{-1}=\begin{pmatrix}q\_{i,j}\end{pmatrix}\_{n\times n}$ is given by
$$
\begin{cases}
q\_{i,j}=0,&1\le j<i\le n;\\
q\_{i,i}=\dfrac{1}{i},&1\le j=i\le n;\\ 
q\_{i,i+1}=-\dfrac{1}{2},&2\le j=i+1\le n;\\
q\_{i,i+k}=u\_{k+1}(i+1)\cdots(i+k-1),&k+1\le i+k\le n, k\ge 2.
\end{cases}
$$
The proof is formal and should be direct, also the sequence $u\_k$ has some zero entries.


