Divisor function asymptotics Define $\tau_{r}(n) = \sum_{d_1...d_r = n}1$. 
One exercise in a book on sieve theory asked for an elementary proof by induction of the fact that 
$$\sum_{n\le x}\tau_r(n) = \frac{1}{(r - 1)!}x(\ln x)^{r - 1} + O\left(x(\ln x)^{r - 2}\right)$$
The base case $r = 2$ is easy with reversing the order of summation.
The only other progress that I made was the fact that $\sum_{n\le x}\tau_{r}(n) = \sum_{d\le x}\left\lfloor\frac{x}{d}\right\rfloor\tau_{r- 1}(d)$, but I don't know how to proceed.
 A: We can write
$$\tau_{r+1}(n)= \sum_{d\mid n} \tau_r(d),$$
which leads to your
$$\sum_{n\leqslant x} \tau_{r+1}(n) = \sum_{d\leqslant x} \left\lfloor \frac{x}{d}\right\rfloor \tau_r(d),$$
but we can also write
$$\tau_{r+1}(n) = \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right),$$
and that gives us
$$\begin{align}
\sum_{n\leqslant x} \tau_{r+1}(n)
&= \sum_{n\leqslant x} \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right)\\
&= \sum_{d\leqslant x} \sum_{k\leqslant \frac{x}{d}} \tau_r(k)\\
&= \sum_{d\leqslant x} C_r\frac{x}{d}\left(\ln \frac{x}{d}\right)^{r-1} + O\left(\frac{x}{d}\left(\ln \frac{x}{d}\right)^{r-2}\right).
\end{align}$$
Now we can estimate the lower-order part by $K\cdot \frac{x}{d}(\ln x)^{r-2}$, and since $\sum_{d\leqslant x} \frac{1}{d} = \ln x + O(1)$, the sum of these terms is, as it should be, $O(x(\ln x)^{r-1})$.
For the dominant terms, we find
$$\begin{align}
\sum_{d\leqslant x} \frac{x}{d}\left(\ln \frac{x}{d}\right)^{r-1}
&= \sum_{d\leqslant x} \frac{x}{d}\left(\ln x - \ln d\right)^{r-1}\\
&= \sum_{d\leqslant x} \frac{x}{d} \sum_{k=0}^{r-1} (-1)^k\binom{r-1}{k} (\ln x)^{r-1-k}(\ln d)^k\\
&= \sum_{k=0}^{r-1}(-1)^k x(\ln x)^{r-1-k}\binom{r-1}{k} \sum_{d\leqslant x} \frac{(\ln d)^k}{d}\\
&= \sum_{k=0}^{r-1}(-1)^k x(\ln x)^{r-1-k}\binom{r-1}{k} \left(\frac{(\ln x)^{k+1}}{k+1} + O(1)\right)\\
&= x(\ln x)^r\sum_{k=0}^{r-1}(-1)^k\binom{r-1}{k}\frac{1}{k+1} + O\left(x(\ln x)^{r-1}\right),
\end{align}$$
and since
$$\sum_{k=0}^{r-1} (-1)^k\binom{r-1}{k}\frac{1}{k+1} = -\frac{1}{r} \sum_{k=0}^{r-1} (-1)^{k+1}\binom{r}{k+1} = \frac{1}{r},$$
we get
$$\sum_{n\leqslant x} \tau_{r+1}(n) = \frac{C_r}{r} x(\ln x)^r + O\left(x(\ln x)^{r-1}\right).$$
Since $\tau_1(n) = 1$ for all $n$, we immediately have
$$\sum_{n\leqslant x} \tau_1(n) = \lfloor x\rfloor = 1\cdot x(\ln x)^0 + O\left(x(\ln x)^{-1}\right),$$
hence $C_1 = 1 = \frac{1}{0!}$, and the recurrence $C_{r+1} = C_r/r$ yields $C_r = \frac{1}{(r-1)!}$, as desired.
