Take a look at this: $$\int_0^t\Gamma(x+1)dx=\sum_{k=0}^\infty\frac{(-t)^k\Gamma^{(k)}(t)}{(k+1)!}\tag{1}$$($\Gamma$ is the analytic continuation of $(x-1)!$ and $f^{(k)}(x)$ means the $k$th derivative of $f$). It is very close to the expansion of $e^{-t}-1$ but with a $\Gamma^{(k)}(t)$ twist. Please post other answers (maybe new equivalences, since the above sum is really just a manipulation of the Taylor series), since the solution here isn't really interesting and I can't figure out another one (It took me two non-continuous months to find the answer when I remembered a nifty integration technique).
Solution: $$\begin{align}\int_0^t\Gamma(x+1)dx&=x\Gamma(x+1)\bigg|_0^t-\int_0^tx\Gamma'(x+1)dx\\ \\ &=t\Gamma(t+1)-\frac{x^2}2\Gamma'(x+1)\bigg|_0^t+\int_0^t\frac{x^2}2\Gamma''(x+1)dx\end{align}$$
This is done by repeated integration by parts. Induction gives $(1)$.
In fact there's nothing special about $\Gamma$ here. This is just the Taylor series about $t$ of an antiderivative of $\Gamma$, evaluated at $0$.