Let's indicate your matrix as
$$
{\bf A}_{\,h} = \left\| {\;a_{\,n,\,m} \quad \left| {\;0 \le n \le m \le h} \right.\;} \right\| = \left\| {\;{{\left[ {n \le m} \right]} \over {m + 1}}\;} \right\|_{\,h}
$$
so that it is a $(h+1) \times (h+1)$ matrix, indexed from $0$, and where the square brackets indicate the Iverson's bracket.
Then let's define
$$
\eqalign{
& {\bf E}_{\,h} = \left\| {\;\left[ {n + 1 = m} \right]\;} \right\|_{\,h} \cr
& {\bf S}_{\,h} = \left\| {\;\left[ {n \le m} \right]\;} \right\|_{\,h} = {{{\bf I}_{\,h} } \over {{\bf I}_{\,h} - {\bf E}_{\,h} }}\quad \cr
& \left( {f(n) \circ {\bf I}_{\,h} } \right) = \left\| {\;f(n)\left[ {n = m} \right]\;} \right\|_{\,h} \cr}
$$
so that
${\bf E}_{\,h}$ is the Shifting matrix, with $1$ on the first upper sub-diagonal and null otherwise;
${\bf S}_{\,h}$ is the Summing matrix, upper triangular with all $1$ 's;
$\left( {f(n) \circ {\bf I}_{\,h} } \right)$ is the diagonal matrix, with diagonal elements $f(0),f(1), \cdots, f(h)$.
Then, as you correctly pointed out, we can write:
$$
{\bf A}_{\,h} = {\bf S}_{\,h} \left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right)
$$
which means
$$
{\bf A}_{\,h} ^{\; - \,{\bf 1}} = \left( {\left( {n + 1} \right) \circ {\bf I}_{\,h} } \right)\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)
$$
It is not difficult to demonstrate that
$$
\eqalign{
& \left( {f(n) \circ {\bf I}_{\,h} } \right)\;{\bf E}_{\,h} = {\bf E}_{\,h} \;\left( {f(n - 1) \circ {\bf I}_{\,h} } \right) = \cr
& = \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {f(k)} \circ {\bf I}_{\,h} } \right)^{\; - \,{\bf 1}} \;{\bf E}_{\,h} \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {f(k)} \circ {\bf I}_{\,h} } \right) \cr}
$$
and that, given in general
$$
{\bf A}_{\,h} (f) = {\bf S}_{\,h} \left( {f(n) \circ {\bf I}_{\,h} } \right)\quad \left| {\;f(m) \ne f(n)} \right.
$$
the relevant matrix of eigenvectors $\mathbf W$ is :
$$
{\bf W}_{\,h} (f) = \left\| {\;\left[ {n \le m} \right]{{f(m)^{\,m - n} } \over {\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {f(m) - f(k)} \right)} }}\;} \right\|_{\,h} = \left\| {\;\left[ {n \le m} \right]\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {{{f(m)} \over {f(m) - f(k)}}} \right)} \;} \right\|_{\,h}
$$
while the inverse is:
$$
\eqalign{
& {\bf W}_{\,h} ^{\; - \,{\bf 1}} (f)\quad \left| {\;0 \ne f(0)} \right.\quad = \cr
& = \left\| {\;\left[ {n \le m} \right]{{f(m)f(n)^{\,m - n - 1} } \over {\prod\limits_{n + 1\, \le \,k\, \le \,m} {\left( {f(n) - f(k)} \right)} }}\;} \right\|_{\,h} = \left\| {\;\left[ {n \le m} \right]{{f(m)} \over {f(n)}}\prod\limits_{n + 1\, \le \,k\, \le \,m} {{{f(n)} \over {\left( {f(n) - f(k)} \right)}}} \;} \right\|_{\,h} \cr}
$$
Now, in the present case, we obtain
$$
\eqalign{
& {\bf W}_{\,h} = \left\| {\;\left[ {n \le m} \right]\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {{{k + 1} \over {k - m}}} \right)} \;} \right\|_{\,h} = \cr
& = \left\| {\;\left[ {n \le m} \right]{{m!} \over {n!}}\prod\limits_{1\, \le \,j\, \le \,m - n} {\left( { - {1 \over j}} \right)} \;} \right\|_{\,h} = \left\| {\;\left( { - 1} \right)^{\,m - n} \left( \matrix{
m \cr
n \cr} \right)\;} \right\|_{\,h} = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \cr}
$$
where $ \overline {\bf B} _{\,h} $ is the transpose of the Lower Triangular Pascal array.
Finally
$$
{\bf A}_{\,h} = {\bf S}_{\,h} \left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right) = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \;\left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right)\;\overline {\bf B} _{\,h}
$$
which demonstrates your assumption.
Concerning your last question instead, we have of course that
$$
{\bf A}_{\,h} ^{\;{\bf q}} = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \;\left( {{1 \over {\left( {n + 1} \right)^{\;q} }} \circ {\bf I}_{\,h} } \right)\;\overline {\bf B} _{\,h}
$$
whose single components are:
$$
\eqalign{
& a^{\left( q \right)} _{\,n,\,m} = \sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{
k \cr
n \cr} \right){1 \over {\left( {k + 1} \right)^{\;q} }}\left( \matrix{
m \cr
k \cr} \right)} = \cr
& = \sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{
m \cr
n \cr} \right)\left( \matrix{
m - n \cr
k - n \cr} \right)\left( {k + 1} \right)^{\; - \,q} } = \cr
& = \left( \matrix{
m \cr
n \cr} \right)\sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{
m - n \cr
k - n \cr} \right)\left( {k - n + n + 1} \right)^{\; - \,q} } = \cr
& = {{m!} \over {n!}}\sum\limits_{\,0\, \le \,j\, \le \,m - n} {{{\left( { - 1} \right)^{\,j} } \over {j!\left( {m - n - j} \right)!}}\left( {j + n + 1} \right)^{\; - \,q} } \cr}
$$
Cannot find at the moment a "more compact" expression.
