Truncating the infinite series for the derivative of the Digamma function $$ \psi'(x) = \sum_{n=0}^\infty\frac{1}{(x + n)^2} $$ after $m-1$ terms, where $m$ is a positive integer (the case $m=2$ answered the question How do we prove that $(x-1)!\leq{(\frac{x}{2})^{x-1}}$?), finding upper and lower bounds for the remainder, and integrating twice between the limits $2$ and $2+x$ (at least, I think that's what I did, but it was a long slog, and my notation has changed several times since then), one arrives at the inequalities $$ \left(\frac{m+1+x}{m+1}\right)^{m+1+x} \!\!\! < \frac{e^x(m+x)!}{e^{(H_m-\gamma)x}m!} < \left(\frac{m+x}{m}\right)^{m+x} \quad (x > 0;\ m = 1, 2, 3, \ldots). $$ This seems most useful (if useful at all!) for smallish $x$. Replacing $m+x$ by $x$ and $m$ by $\left\lfloor{x}\right\rfloor$, we get $$ \left(\frac{x+1}{\left\lfloor{x}\right\rfloor+1}\right)^{x+1} \!\!\! < \frac{e^{x-\left\lfloor{x}\right\rfloor}}{e^{(H_m-\gamma)(x-\left\lfloor{x}\right\rfloor)}} \cdot \frac{x!}{\left\lfloor{x}\right\rfloor!} < \left(\frac{x}{\left\lfloor{x}\right\rfloor}\right)^x \quad(x > 1,\ x \notin \mathbb{N}). $$ This seems to give sharper bounds than the following simple exact form of Stirling's approximation: $$ \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n} \leqslant n! \leqslant en^{n+\frac{1}{2}}e^{-n}. $$ On the other hand, it seems to be generally inferior to the full version of Robbins's bounds: $$ \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n+1}} < n! < \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n}}. $$ For small values of $x - \left\lfloor{x}\right\rfloor$, my formula does sometimes give better results. For example, $7.04! \bumpeq 5463.7647$, and in this case my formula gives the strict bounds $(5463.7292, 5463.8071)$, whereas Robbins's formula gives $(5463.0514, 5463.8080)$, and the simplified version of his formula gives the distinctly worse estimates $(5399.5135, 5855.4353)$.
Might my horrid formula therefore have some actual use? If so, has it been published already? Does it have a less nasty proof than the one I've sketched?