I am using the textbook Probability and Measure, Anniversary edition by Patrick Billingsley, this is exercise 1.6.
Put $M(t) = \int_{0}^{1}e^{t s_n(\omega)}d\omega$ and show by successive differentiation's under the integral that \begin{equation} M^{(k)}(0) = \int_{0}^{1}s_{n}^{k}(\omega) d\omega. \end{equation} Over each dyadic interval of rank $n$, $s_n(\omega)$ has a constant value of the form $\pm 1 \pm 1 \pm \dots \pm 1$. Therefore $M(t) = 2^{-n}\sum \exp t(\pm 1 \pm 1 \pm \dots \pm 1$). Thus \begin{equation} M(t) = \bigg(\frac{e^t + e^{-t}}{2} \bigg)^n = (\cosh t)^n \end{equation} Using these two equalities, we are derive the following results: $$\int_{0}^{1}s_n(\omega) d\omega = 0 $$ $$\int_{0}^{1}s_{n}^{2}d\omega = n$$ $$\int_{0}^{1}s_{n}^{4}(\omega)d\omega = n + 3n(n-1) \leq 3n^2 $$
Note that $s_n$ is defined as the partial sum of the $n$ Rademacher functions.
The first equality makes sense to me. Set $t=0$ and by induction, the equality holds. How does the moment generating function equal the hyperbolic function $(\cosh t)^n$ in the second equality? I don't understand intuitively what is going on here.
I think that I am going in the wrong direction with solving the problem and would just like a hint (NOT SOLUTIONS) to get me started as I am out of ideas right now, thanks