# help with an asymptotic for a certain product

I'm having difficulty finding an asymptotic formula for the following product:

$$k^{\alpha}\prod_{1 \leq i \leq N \atop i \neq k} (i^{\alpha}-k^{\alpha})$$ where $N$ is a parameter tending to infinity and $\alpha = \alpha(N)$ is on the order of $N^{-1}$.

The method I've tried to use to estimate this product is by giving a second order Taylor expansion of $e^{\alpha \log i} - e^{\alpha\log k}$ with $1 \leq i, k \leq N$, and splitting the product into the ranges $1 \leq i \leq k-1$ and $k+1 \leq i \leq N$ (a higher order expansion seems unnecessary given how small $\alpha$ is and how many terms are included in the summation). Splitting the product in this fashion (after factoring out $(-1)^{k-1}\alpha^{N-1}$ and making some simpler estimates) gives rise to two products \begin{align*} \prod_{1 \leq i \leq k-1} \log(k/i) = \exp\left(\sum_{1 \leq i \leq k-1} \log_2(k/i)\right) \\ \prod_{k+1\leq i \leq N} \log(i/k) = \exp\left(\sum_{k+1 \leq i \leq N} \log_2(i/k)\right), \end{align*} (here $\log_2 t := \log(\log t)$) and it seems that Euler-Maclaurin summation is the obvious move here in order to estimate the arguments of the exponents (I hope to get, at the very least, $o(1)$ error terms for these arguments as $N \rightarrow \infty$). On the other hand, the summands in the first and second integral blow up whenever $k/i$ or $i/k$ is very close to 1 (which will occur when $N$ gets large since $k$ is meant to be generic), respectively; if we ignore terms of the form $i/k \in ((1+\delta)^{-1},1+\delta)$, where $\delta > 0$ is some parameter that we can choose then we get sums that are more or less computable via Euler-Maclaurin if $\delta$ is not chosen too small (since the integral $\int_{1+\delta}^{N/k} \log_2 t \ dt$ can be computed in relation to the logarithmic integral for example, which has well-studied asymptotic expansions).
The sum over the $\delta$-interval that we've omitted, though, is difficult to compute for general values of $k$, and it seems to me that some upper bound must be taken: for instance, i we use the trivial bound (e.g., for $k+1 \leq i \leq (1+\delta)k$ we take $|\log_2 (i/k)| \leq |\log_2(1+1/k)| \leq 2\log k$) then this middle sum gives about $\delta k \log k$ as a contribution, and clearly, then, $\delta$ must be chosen very small in order to give an error term in the exponential that gives rise to an asymptotic formula, and this is unsuitable.

I would appreciate any sort of a hint (but, please, no full solutions) to answer at least one of the following questions:

1. is there a better approach to this estimate in order to get a small error term without exponentiation?

1. Is there a better technique to bound the sum over the $((1+\delta)^{-1},1+\delta)$ interval than what is presented above?

Any input is sincerely appreciated.