Is it true that for all $k>1$, $ x^k > x \log x $ for sufficiently large $x$?

If $k > 1$, then $k = 1 + \varepsilon$ for some $\varepsilon > 0$. So $x^k = x^{1 + \varepsilon} = xx^\varepsilon$.

Thus, to show $x^k = xx^\varepsilon > x\log(x)$, eventually, amounts to showing $x^\varepsilon > \log(x)$, eventually. And, this latter fact is true because: \begin{align*} \lim_{x \to \infty}\frac{\log(x)}{x^\varepsilon} &= \lim_{x \to \infty} \frac{1/x}{\varepsilon x^{\varepsilon - 1}}\quad \text{ (L'Hôpital's rule)} \\ &= \lim_{x \to \infty} \frac{1}{\varepsilon x^\varepsilon} = 0\quad \text{ ($x^\varepsilon \to \infty$ as $\varepsilon > 0$)} \end{align*}

Answer to Edit:

The reason you have a crossing point converging to $e$ as $k \to 1$ is quite simple. Just let $k = 1$ and let us calculate the crossing points. Crossing points occur when: $x^k = x = x \log(x)$. Or rearranging: $$ x - x\log(x) = x (1 - \log(x)) = 0 $$ If you ignore $x = 0$, then the above is true when $1 - \log(x) = 0$ or $\log(x) = 1$ or $x = e$.