Simple equivalent of $\int_0^\infty\frac{dx}{(x+1)(x+2)...(x+n)}$ when $n\to\infty$

Note that $$ \begin{align} &\int_0^\infty\frac{(n-1)\,\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}\\ &=\int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)} -\int_0^\infty\frac{\mathrm{d}x}{(x+2)(x+3)\cdots(x+n)}\\ &=\int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)} -\int_1^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}\\ &=\int_0^1\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}\tag{1} \end{align} $$ Therefore, $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)} &=\frac1{n-1}\int_0^1\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}\\ &=\frac1{n-1}\int_0^1\frac{\Gamma(x+1)}{\Gamma(x+n)}\mathrm{d}x\tag{2} \end{align} $$ For $x\in[0,1]$, Gautschi's Inequalty says that $\frac{n^{1-x}}{n!}\le\frac1{\Gamma(x+n)}\le\frac{(n+1)^{1-x}}{n!}$. Furthermore, $1-\gamma\,x\le\Gamma(x+1)\le1$ for $x\in[0,1]$. Thus, $(2)$ is greater than or equal to $$ \frac1{(n-1)n!}\int_0^1(1-\gamma\,x)n^{1-x}\,\mathrm{d}x=\frac1{n!\log(n)}\left(1-\gamma\left(\frac1{\log(n)}-\frac1{n-1}\right)\right)\tag{3} $$ and less than or equal to $$ \frac1{(n-1)n!}\int_0^1(n+1)^{1-x}\,\mathrm{d}x=\frac{n}{n-1}\frac1{n!\log(n+1)}\tag{4} $$ Therefore, $$ \int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}\sim\frac1{n!\log(n)}\tag{5} $$

Let $I_n$ denote the $n$th integral, then the lower bound $1/(1+u)\geqslant\mathrm e^{-u}$, valid for every $u\geqslant0$, yields $$ \frac1{(x+1)\cdots(x+n)}=\frac1{n!}\prod_{k=1}^n\frac1{1+\frac{x}k}\geqslant\frac1{n!}\mathrm e^{-xH_n}, $$ where $H_n=\sum\limits_{k=1}^n\frac1k$ denotes the $n$th harmonic number, hence $H_n\sim\log n$. Integrating this, one gets, for every $n\geqslant2$, $$ I_n\geqslant \frac1{n!H_n}, $$ and I would be inclined to think, but did not prove at the moment, that $1/(n!\log n)$ is also the correct equivalent of $I_n$ when $n\to\infty$.

The best I can prove at the moment uses the lower bound $(x+1)\cdots(x+n)\geqslant\frac12n!(x+1)(x+2)$, valid for every $x\geqslant0$. Thus, for every $n\geqslant2$, $$ I_n\leqslant\frac2{n!}I_2=\frac{2\log2}{n!}. $$

$Edit: Thank to @Did and @Robjohn I have succeeded: I am going to use Lebesgue's Dominated Convergence Theorem.

Let In denote the nth integral and $H_n=\sum_{k=1}^n\frac{1}{k}$,

$$I_n=\frac{1}{n!}\displaystyle\int_0^\infty\frac{1}{(1+x)(1+\frac{x}{2})...(1+\frac{x}{n})}dx=\frac{1}{n!H_n}\displaystyle\int_0^\infty\frac{1}{(1+\frac{u}{H_n})...(1+\frac{u}{nH_n})}dx$$ with $u=xH_n$

Then we have $(1+\frac{u}{H_n})...(1+\frac{u}{nH_n})\xrightarrow[n\to+\infty]. e^u\tag{1}$

Proof: $$ln((1+\frac{u}{H_n})...(1+\frac{u}{nH_n}))=\sum_{k=1}^n ln(1+\frac{u}{kH_n})$$ Yet we know that : $\forall u\in[0,1], u-\frac{u^2}{2}\leq ln(1+u)\leq u$ $$\sum_{k=1}^n(\frac{u}{kH_n}-\frac{u^2}{2k^2H_n^2})\leq\sum_{k=1}^n ln(1+\frac{u}{kH_n})\leq \frac{u}{H_n}\sum_{k=1}^n \frac{1}{k}$$ $$u-\frac{u^2}{2H_n^2}\sum_{k=1}^n\frac{1}{k^2}\leq\sum_{k=1}^n ln(1+\frac{u}{kH_n})\leq u$$

Furthermore, the series $\sum\frac{1}{k^2}< \infty$, so $\sum_{k=1}^\infty\frac{1}{k^2}=S$

Therefore, $$e^u\leq \lim_{n\rightarrow\infty} \prod_{k=1}^n (1+\frac{u}{kH_n})\leq e^u$$ QED

The last step: $$(1+\frac{u}{H_n})...(1+\frac{u}{nH_n})\geq 1+u+\frac{u^2}{H_n^2}(\sum_{1\leq i<j\leq n} \frac1{ij})$$ $$=1+u+\frac{u^2}{H_n^2}(H_n^2-\sum_{1}^n\frac1{i^2})$$ $$\geq 1+u+u^2(1-\frac{\pi^2}{6H_n^2})$$ $$\geq 1+u+\frac{u^2}{2}\tag{2}$$

(1)+(2)+Lebesgue's Dominated Convergence Theorem $\Longrightarrow \int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}\sim\frac1{n!\ln(n)}$