Just for fun, I defined the following series. For $m\in\Bbb N$, 
$$g\_m=\sum\_{n\geq1}\prod\_{i=0}^{m}\frac1{n+i}$$
I am seeking a closed form for $g\_m$. I found $g\_1,\ g\_2$, and a general expression involving a lot of integrals.

$$
\begin{align}
g\_1=&\sum\_{n\geq1}\frac1{n(n+1)}\\
=&\sum\_{n\geq1}\frac1n\int\_0^1 t\_0^n\mathrm{d}t\_0\\
=&\sum\_{n\geq1}\int\_0^1\int\_0^{t\_0} t\_1^{n-1}\mathrm{d}t\_1\mathrm{d}t\_0 \\
=&\int\_0^1\int\_0^{t\_0}\sum\_{n\geq0}t\_1^{n}\mathrm{d}t\_1\mathrm{d}t\_0 \\
=&\int\_0^1\int\_0^{t\_0}\frac{\mathrm{d}t\_1}{1-t\_1}\mathrm{d}t\_0 \\
=&1 \\
\end{align}
$$
Through the same process, it can be shown that $g\_2=\frac14$. In fact, through the same process, it can be shown that 
$$g\_m=\int\_0^1\int\_0^{t\_0}\int\_0^{t\_1}\cdots\int\_0^{t\_{m-2}}\frac{\mathrm{d}t\_{m-1}}{1-t\_{m-1}}\mathrm{d}t\_{m-2}\cdots\mathrm{d}t\_2\mathrm{d}t\_1\mathrm{d}t\_0$$
Which I don't even know how to approach. Any ideas? Thanks.

**This isn't a duplicate question, because I would like to know how to evaluate those nested integrals, if possible.**