0, \ldots, d$.

Consider the function $g_d$ defined by $$g_d(t) := \frac{2 (1 - e^t)^{d + 1}}{e^t + 1}
    = \sum_{\ell = 0}^{d + 1} (-1)^\ell \binom{d+1}{\ell}
      \frac{2 e^{\ell t}}{e^t + 1}.$$ From (\[5.4\]) with $x = \ell$, the Taylor expansion of $g_d$ can be written as $$g_d(t)
    = \sum_{n = 0}^\infty \left[
      \sum_{\ell = 0}^{d + 1} (-1)^\ell \binom{d + 1}{\ell} E_n(\ell)
      \right] \frac{t^n}{n!},$$ so $$D^n g_d(t) \big|_{t = 0}
    = \sum_{\ell = 0}^{d + 1} (-1)^\ell \binom{d + 1}{\ell} E_n(\ell),
    \quad n \geq 0.$$ On the other hand, the Leibniz rule gives $$D^n g_d(t) = \sum_{m = 0}^{n} \binom{n}{m}
    D^m \left[ (1 - e^t)^{d + 1} \right] \,
    D^{n - m} \left[ \frac{2}{e^t + 1} \right].$$ If $n \leq d$, then $D^m \left[ (1 - e^t)^{d + 1} \right]$ is always divisible by $(1 - e^t)$. Hence, $$D^n g_d(t) \big|_{t = 0} = 0, \quad n = 0, \ldots, d,$$ which proves the first identity (\[5.7\]).

The second one follows similarly, by considering $$h_d(t) := g_d(t) - g_{d + 1}(t) = \frac{2 e^t (1 - e^t)^{d + 1}}{e^t + 1}
    = \sum_{\ell = 0}^{d + 1} (-1)^\ell \binom{d+1}{\ell}
    \frac{2 e^{(\ell + 1) t}}{e^t + 1}.$$ The Taylor expansion 