We define $\tau_{k}\left(n\right)$ to be the number of ordered $k$-tuples of positive integers with product equal to $n$. It is easily shown that this satisfies the recurrence relation $\tau_{1}\left(n\right)=1$ and $\tau_{k}\left(n\right)=\sum_{d\mid n}\tau_{k-1}\left(d\right)$, and I would like to show, using $\log^{k}x$ to denote $\left(\log x\right)^{k}$, that $$\sum_{n\leq x}\tau_{k}\left(n\right)=O\left(x\log^{k-1}x\right).$$
The context for this problem essentially is the same as in my previous question. In the paper by Hu, he uses $\sum_{n\leq x}\frac{\tau_{k}\left(n\right)}{n}=O\left(\log^{k}x\right)$, citing this paper by László Tóth (PDF file). In that paper, Tóth uses $\sum_{n\leq x}\tau_{k}\left(n\right)=O\left(x\log^{k-1}x\right)$, which he calls a "familiar result". Since it is not so familiar to me, I thought I'd take a shot at proving it.
I proceed by induction. For $k=1$, we have $\sum_{n\leq x}\tau_{1}\left(n\right)=\sum_{n\leq x}1=O\left(x\right).$
Now, suppose that $\sum_{n\leq x}\tau_{k}\left(n\right)=O\left(x\log^{k-1}x\right).$ Then, we have
$$\sum_{n\leq x}\tau_{k+1}\left(n\right)=\sum_{n\leq x}\sum_{d\mid n}\tau_{k}\left(d\right)=\sum_{ab\leq x}\tau_{k}\left(b\right)=\\ \sum_{a\leq x}\sum_{b\leq x/a}\tau_{k}\left(b\right)=\sum_{a\leq x}O\left(\frac{x}{a}\log^{k-1}\frac{x}{a}\right)=\\ O\left(x\log^{k-1}x\sum_{a\leq x}\frac{1}{a}\right)=O\left(x\log^{k}x\right).\textbf{ QED}(?)$$
Is my attempt at a proof correct? My main issue is the step $\sum_{a\leq x}O\left(\frac{x}{a}\log^{k-1}\frac{x}{a}\right)=O\left(x\log^{k-1}x\sum_{a\leq x}\frac{1}{a}\right)$, as I am unsure if this is allowed. If this turns out to be a faulty step, how would I go about proving this result?
Thank you for your time and help.
Use partial summation formula:
$$\sum_{n\leq x}\sum_{d\mid n}\tau_{k}\left(d\right) = \sum_{d \le x} \tau_k(d)\left[\frac{x}{d}\right] = \mathcal{O}\left(x\int_1^x \frac{\log^{k-1} t}{t}\,dt\right) = \mathcal{O}\left(x\log^k x\right)$$