Showing $[K(x):K(\frac{x^5}{1+x})]=5$?

We only need to show that the polynomial $x^5-\alpha x - \alpha$ is irreducible over $K(\alpha)[x]$. Note that the polynomial is primitive, thus it suffices to show it is irreducible over $K[\alpha][x]$.

We use the idea of Eisenstein criterion, let $\varphi:K[\alpha]\to K$ be the evaluation map sending $\alpha \mapsto 0$. Extend this homomorphism to get $\pi: K[\alpha][x]\to K[x]$. If $x^5-\alpha x -\alpha = fg$, with $f,g \in K[\alpha][x]$, $f,g$ non-unit. Then $$\pi(f)\pi(g) = x^5$$ Note that $\ker \pi = (\alpha)$, the ideal generated by $\alpha$.

Since $K[x]$ is a UFD, $f = x^r + \alpha f_0, g = x^{5-r} + \alpha g_0$, where $f_0,g_0\in K[\alpha][x]$. But this would imply that the constant term of $fg$ is a multiple of $\alpha^2$. Hence $x^5-\alpha x -\alpha$ is irreducible over $K[\alpha][x]$.