Is this a valid identity?
$$\binom{m+n}{m}=\frac{(m+n)!}{m!n!}$$ $$=\frac{m+n}{n}\cdot\frac{m+(n-1)}{n-1}\cdot\frac{m+(n-2)}{n-2}\cdots\frac{m+2}{2}\cdot\frac{m+1}{1}$$ $$\left(1+\frac{m}{1}\right)\cdot\left(1+\frac{m}{2}\right)\cdot\left(1+\frac{m}{3}\right)\cdots\left(1+\frac{m}{n}\right)\cdot$$ Expanding here gives me a $m$-th degree polynomial in $m$ $$1+\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n}\right)m+\left(\frac{1}{1\cdot 2}+\frac{1}{1\cdot 3}+\cdots+\frac{1}{(n-1)\cdot n}\right)m^2+\cdots+\frac{1}{n!}m^n$$ $$=\sum_{k=0}^n H_{n,k}m^k$$
From the Taylor expansion of \begin{align} \binom{x+n}{n} = \sum_{k=0}^{\infty} a_{k}(n) \, x^{k} \end{align} it is found that \begin{align} f(x) &= \binom{x+n}{n} \\ f'(x) &= \binom{x+n}{n} \left[ \psi(x+n+1) - \psi(x+1) \right] \\ f''(x) &= \binom{x+n}{n} \left[ \psi'(x+n+1) - \psi'(x+1) + \left(\psi(x+n+1) - \psi(x+1) \right)^{2} \right] \\ f'''(x) &= \binom{x+n}{n} \left[ \psi''(x+n+1) - \psi''(x+1) + 3 \left(\psi(x+n+1) - \psi(x+1) \right) \left(\psi'(x+n+1) - \psi'(x+1) \right) \right. \\ & \hspace{25mm} \left. + \left(\psi(x+n+1) - \psi(x+1) \right)^{3} \right] \end{align} for which \begin{align} f(0) &= 1 \\ f'(0) &= H_{n} \\ f''(0) &= \psi'(n+1) - \psi'(1) + H_{n}^{2} \\ f'''(0) &= \psi''(n+1) - \psi''(1) + 3 H_{n} \, \left(\psi'(n+1) - \psi'(1) \right) + H_{n}^{3} \end{align} This leads to \begin{align} \binom{x+n}{n} &= 1 + H_{n} \, x + \left( \psi'(n+1) - \psi'(1) + H_{n}^{2} \right) \, x^{2} \\ & \hspace{10mm} + \left( \psi''(n+1) - \psi''(1) + 3 H_{n} \, \left(\psi'(n+1) - \psi'(1) \right) + H_{n}^{3} \right) \, x^{3} + \cdots \end{align}