Appearance of log(2) in attempts to solve the Erdős-Moser equation Let $m \geq 2$ be an integer, and denote by $k(m) > 0$ the unique positive real number such that
$$\sum_{j=1}^{m-1} j^{k(m)} = 1^{k(m)} + 2^{k(m)} + \ldots + (m-1)^{k(m)} = m^{k(m)}$$
holds.
Question 1: What is the most elementary way to prove that
$$\lim_{m \to \infty} \frac{k(m)}{m} = \log(2)?$$
Question 2: Assume that we know (for whatever reason) that $\frac{k(m)}{m} = c + O(\frac{1}{m})$ as $m \to \infty$, where $c \in \mathbb{R}$ is some constant. (Indeed, this is true, e.g. by Theorem 1 in   arXiv:0907.1356.) Is there an easy way to see that $c = \log(2)$?
I'm asking because I'm currently planning a course for undergraduate students on continued fractions and Diophantine approximation. I really would like to include the above result, but all proofs that I've found so far seem to be quite involved. Any help is appreciated!
 A: I would like to share here how I finally explained the appearance of $\log(2)$ in my undergraduate course:
For any integer $m > 3$ we shall denote by $k(m) > 1$ the unique real number such that
$$1^{k(m)} + \ldots + (m-1)^{k(m)} = m^{k(m)}$$
holds. Then we observe that
$$m^{k(m)} \leq \int_0^m x^{k(m)} \text{ } \mathrm{d}x = \frac{m^{k(m)+1}}{k(m)+1}$$
and
$$2 m^{k(m)} = 1^{k(m)} + \ldots + (m-1)^{k(m)} + m^{k(m)} \geq \int_0^m x^{k(m)} \text{ } \mathrm{d}x = \frac{m^{k(m)+1}}{k(m)+1},$$
hence $k(m)+1 \leq m \leq 2(k(m)+1)$, proving that the sequence $\big( \frac{k(m)}{m} \big)_{m > 3}$ is bounded by non-zero constants from above and below.
Given this, we may apply Théorème 2 from this paper by H. Delange to obtain that
$$m^{k(m)} = \sum_{r=1}^{m-1} r^{k(m)} = \frac{(m-1)^{k(m)}}{1-e^{-(k(m)+1)/(m-1)}} \Big( 1 + \mathcal{O}\big( \frac{1}{\sqrt{k(m)}} \big)\Big) \text{ as } m \to \infty.$$
(By the way, the proof of the latter theorem seems to be a nice application of the residue theorem from complex analysis.)
This may be rephrased as
$$1 = \frac{e^{- \frac{k(m)}{m} + \mathcal{O}(\frac{1}{m})}}{1 - e^{- \frac{k(m)}{m} + \mathcal{O}(\frac{1}{m})}} \cdot \Big( 1 + \mathcal{O} \big( \frac{1}{\sqrt{k(m)}} \big) \Big) \text{ as } m \to \infty,$$
where we used that
$$\Big( 1 - \frac{1}{m} \Big)^m = e^{-1 + \mathcal{O}(1/m)} \text{ as } m \to \infty$$
and (once again) that $\big( \frac{k(m)}{m} \big)_{m > 3}$ is bounded. After multiplication by $1 - e^{- \frac{k(m)}{m} + \mathcal{O}(\frac{1}{m})}$, we thus have
$$1 = 2 \cdot e^{- \frac{k(m)}{m} + \mathcal{O}(\frac{1}{m})} + e^{- \frac{k(m)}{m} + \mathcal{O}(\frac{1}{m})} \cdot \mathcal{O} \big( \frac{1}{\sqrt{k(m)}} \big) \text{ as } m \to \infty.$$
But
$$e^{- \frac{k(m)}{m} + \mathcal{O}(\frac{1}{m})} = \mathcal{O}(1) \text{ and } e^{\mathcal{O}(\frac{1}{m})} = 1 + o(1) \text{ as } m \to \infty,$$
so that we end up with
$$1 = 2 \cdot e^{- \frac{k(m)}{m}} + o(1) \text{ as } m \to \infty,$$
or in other words,
$$\lim_{m \to \infty} \frac{k(m)}{m} = \log(2).$$
