In the page 14 of the book "The Malliavin Calculus and Related Topics" from Nualart one reads:
I understand that if $Z_t \sim N(0,t)$ then
$$\Bbb{E}[Z_t^{2k}] = \frac{(2k)!}{2^k k!}t^k$$
However, I can't see why this follows from the single fact that $\Bbb{E}[(W_t-W_s)^2] = t-s$
For each $i$, he defines $\{W^i(t), t \in \mathbb{R}^+\}$ to be a zero-mean Gaussian process with covariance function $E[W^i(s)W^i(t)] = s \wedge t$. This means that we know that $W^i(t) - W^i(s)$ is a normal random variable with mean zero; therefore, once we know what it's variance is, we've characterized the distribution entirely. The point of this section is to use Kolmogorov's continuity criterion to show that we can make $W^i$ continuous almost surely.
EDIT: Here's a process that has independent increments, mean zero, and the variance of increments is $t - s$ but isn't Brownian motion:
Consider a Poisson process on $\mathbb{R}^+$ with intensity $1$, and let $X_t$ denote the number of points in $[0,t]$. Define $Z_t = X_t - t$. Then $$\mathbb{E}[(Z_t - Z_s)^2] = \mathbb{E}[((X_t - X_s) - (t-s))^2] = \mathrm{Var}[\mathrm{Poisson}(t-s)] = t-s$$