Is it true that $\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1$? This question is inspired by the formula $$\displaystyle\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots = \zeta(2)-1,$$
see for instance this question.
It appears that for any integer $n\geq 0$, we have
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1.$$
I believe that Yiorgos' answer could be generalized to prove the above formula. Can anyone pull it off?
Side question: If $n$ is replaced by a complex number $s$, for which values of $s$ does the integral converge? (We interpret $x^s$ as $\exp(x\log s)$, where $\log s$ is the principal branch of the complex log.) If the above formula is true, then for any integer $n\geq 0$, have $$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((n+1)x^n-nx^{n-1}) dx = \zeta(n+2) - 1.$$
Is it true that for complex value of $s$ for which the integral converges, we have
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((s+1)x^s-sx^{s-1}) dx = \zeta(s+2)-1?$$
 A: Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ this integral can be rewritten as:
$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(n+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(n+2)}dt \\&= \frac{1}{n+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(n+1)}-(k+1)^{-(n+1)}\right)
\end{align}$$
That's pretty much the exact same as the proof for $n=0$.
Now, 
$$\begin{align}
\frac{1}{k}(k+1)^{-(n+1)}&= \frac{1}{k(k+1)}(k+1)^{-n}\\
&=\left(\frac{1}{k}-\frac{1}{k+1}\right)(k+1)^{-n} \\
&=\frac{1}{k}(k+1)^{-n}- (k+1)^{-(n+1)}\\
&=\dots\text{ induction applied here on }\frac1k(k+1)^{-n}\dots\\
&=\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}-\cdots-\frac{1}{(k+1)^{n+1}}
\end{align}$$
So $$\frac{1}{k}\left(k^{-(n+1)}-(k+1)^{-(n+1)}\right) = \left(\sum_{j=2}^{n+1} \frac{1}{(k+1)^j}\right) + \frac{1}{k^{n+2}}+\frac{1}{k+1}-\frac1k$$
This yields the result you want.
The side question (at least when $s\neq -1$) essentially asks about when:
$$\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(s+1)}-(k+1)^{-(s+1)}\right)$$ converges.  When $s$ is real and $s>-1$, the terms are positive, and bounded above by the sequence $k^{-(s+1)}-(k+1)^{-(s+1)}$, which is a telescoping sequence, so it converges.
A: To answer the side question.
Assume that $ \Re s >1$. Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ the initial integral can be rewritten as:
$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(s+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(s+2)}dt \\&= \frac{1}{s+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(s+1)}-(k+1)^{-(s+1)}\right)\\&= \frac{1}{s+1}\left(\sum_{k=1}^\infty \frac{1}{k^{s+2}}-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}\right)
\end{align}$$
Thus
$$
\int_0^1 \lfloor x^{-1} \rfloor^{-1} (s+1)\:x^s \:\mathrm dx = \zeta(s+2)-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}, \quad  \Re s >1 \tag1
$$ and we have
$$
\int_0^1 \lfloor x^{-1} \rfloor^{-1} s\:x^{s-1} \:\mathrm dx = \zeta(s+1)-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}, \quad  \Re s >2. \tag2
$$
Then we may substract $(2)$ from $(1)$, using
$$
\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}=\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}\left(1-\frac{1}{(k+1)}\right)=\zeta(s+1)-1
$$
to get, by analytic continuation, 
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((s+1)x^s-sx^{s-1}) dx = \zeta(s+2)-1, \quad \Re s >0.$$
