Compact formula for the $n$th derivative of $f\left(\sqrt{x+1}\right)$?

Heuristic

By analyzing the matrix of the first values for $a_{n,m}$ it is possible to deduce that $$ \bbox[lightyellow] { a_{\,n,\,m} =\frac{ \left( { - 1} \right)^{n - m} }{2^n} \left( \matrix{ 2n - 1 - m \cr 2n - 2m \cr} \right) \left( {2n - 2m - 1} \right)!!\quad \left| {\,0 \le m \le n} \right. } \tag{1}$$ which can be further simplified in terms of Gamma or other, and which can be of help for a rigorous demonstration.

Demonstration

We put $$ {{d^{\,n} } \over {dx^{\,n} }}f\left( {\sqrt {x + 1} } \right) = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} {{f^{\,\left( k \right)} \left( {\sqrt {x + 1} } \right)} \over {\left( {x + 1} \right)^{\,n - k/2} }}} $$

The $a_{\,n,\,k} $ are supposed to be constant, independent from $f$.
Then they shall be valid for any infinitely differentiable function, in particular for analytic functions, and thus in particular for $$ f(x) = x^{\,2m} \quad \left| {\,0 \le m \in Z} \right. $$

For this we have $$ \left\{ \matrix{ f\left( {\sqrt {x + 1} } \right) = \left( {x + 1} \right)^{\,m} \hfill \cr {{d^{\,n} } \over {dx^{\,n} }}f\left( {\sqrt {x + 1} } \right) = m^{\,\underline {\,n\,} } \left( {x + 1} \right)^{\,m - n} \hfill \cr f^{\,\left( n \right)} \left( {\sqrt {x + 1} } \right) = \left( {2m} \right)^{\,\underline {\,n\,} } \left( {\sqrt {x + 1} } \right)^{\,2m - n} \hfill \cr} \right. $$ and therefore $$ \eqalign{ & m^{\,\underline {\,n\,} } \left( {x + 1} \right)^{\,m - n} = \cr & = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} {{\left( {2m} \right)^{\,\underline {\,k\,} } \left( {x + 1} \right)^{\,m - k/2} } \over {\left( {x + 1} \right)^{\,n - k/2} }}} = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \left( {2m} \right)^{\,\underline {\,k\,} } \left( {x + 1} \right)^{\,m - n} } \cr & \quad \quad \Downarrow \cr & m^{\,\underline {\,n\,} } = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \left( {2m} \right)^{\,\underline {\,k\,} } } \quad \Leftrightarrow \quad \left( {{m \over 2}} \right)^{\,\underline {\,n\,} } = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} m^{\,\underline {\,k\,} } } \cr} $$

Since $$ \eqalign{ & \left( {{m \over 2}} \right)^{\,\underline {\,n\,} } = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,n - j} \left[ \matrix{ n \cr j \cr} \right]\left( {{m \over 2}} \right)^{\,j} } = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {{{\left( { - 1} \right)^{\,n - j} } \over {2^{\,j} }}\left[ \matrix{ n \cr j \cr} \right]m^{\,j} } \cr & \sum\limits_{0\, \le \,k} {a_{\,n,\,k} m^{\,\underline {\,k\,} } } = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( { - 1} \right)^{\,k - j} \left[ \matrix{k \cr j \cr} \right]m^{\,j} } } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,} {\left( {\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k - j} a_{\,n,\,k} \left[ \matrix{ k \cr j \cr} \right]} } \right)m^{\,j} } \cr} $$ i.e. $$ {{\left( { - 1} \right)^{\,n - j} } \over {2^{\,j} }}\left[ \matrix{ n \cr j \cr} \right] = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k - j} a_{\,n,\,k} \left[ \matrix{ k \cr j \cr} \right]} $$ which can be viewed as a matrix relation involving the matrix $A:\; A_{\,n,\,k} = a_{\,n,\,k}$ multiplied by the matrix of the signed Stirling N. of 1st kind, which is well known to be the inverse of the Stirling N. of 2nd kind, so: $$ \bbox[lightyellow] { a_{\,n,\,m} = \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,n - k} } \over {2^{\,k} }} \left[ \matrix{ n \cr k \cr} \right]\left\{ \matrix{ k \cr m \cr} \right\}} }\tag{2}$$

The computation of the above checks with the heuristic formula:
it remains to demonstrate this fact analytically.