I have that $|a_n - (1+\frac{r}n)| \leq \frac c{n^2}$, for $c$ a constant, and am attempting to show that there exist constants $C < \infty$ and $K > 0$ such that the product $b_n = \prod_{i=1}^n a_i$ can be approximated as $$|b_n - Kn^r| \leq C n^{r-1}.$$
It was suggested that I consider $\log b_n$, and use the Taylor approximation for $\log(1+x)$. However, I have found this approach to be yielding no fruit: while it's certainly true that for sufficiently large $n$, $\log(1+\frac{r}n)$ can be approximated by way of a Taylor series, I can't deal with the additional terms added to this as arguments to $\log$ - and have no idea how to make $r$ appear as an exponent to $n$.
Here are some ideas: first it has to be true for $a_i=1+r/i$, and in this case we can explicitely compute the product $b_n$, which is equal to $\prod_{i=1}^r(1+n/i)$. Taking $K :=1/r!$, we obtain that $\prod_{i=1}^r(1+n/i)-Kn^r$ is of order $n^{r-1}$.
For the general case, we have to control the quantity $$c_n :=\left|b_n-\prod_{i=1}^n \left(1+\frac ri\right) \right|;$$ more precisely, the existence of a constant $C$ such that $c_n\leqslant Cn^{r-1}$ for each $n$. To this aim, define $p_n:=\prod_{i=1}^n \left(1+\frac ri\right)$, and note that $$|b_{n+1}-p_{n+1}|\leqslant \frac{n+1+r}{n+1}|b_n-p_n| +c\frac{b_n}{n^2}\\ \leqslant \left(\frac{n+1+r}{n+1}+\frac c{n^2} \right) |b_n-p_n| +c\frac{p_n}{n^2}. $$ From this, I think that you can prove by induction that a $C$ which works for $n=1$ actually does the job for each $n$.