Let $f_n(t)=\frac{n!}{(t+1)(t+2)\cdots (t+n+1)}$ prove that $\int_0^1 f_n(t)dt\sim \frac{1}{n\log n}$ when $n\to\infty$.
I've noticed that $f_n(t)=\frac{\Gamma(n+1)\Gamma(t+1)}{\Gamma(t+n+2)}=\int_0^1dt\int_0^1 x^t(1-x)^n dx$
then $f_n(t)=\int_0^1(1-x)^n dx\int_0^1 x^t dt=-\int_0^1\frac{(1-x)^{n+1}}{\log x}dx$
but this method seems not work, I don't know how to get the asymptotic.