Let the probability distribution for $X=x$ have a binomial distribution with parameters $n$ and $p$: $$\mathrm{P}(X=x) = \binom{n}{x} p^x\, (1-p)^{n-x}, \quad x \in \{0,\dotsc,n\}.$$
I'm looking for upper and lower bounds of its raw $k$th moment, possibly as functions of $k$, $n$, $p$: $$L(k,n,p) \leqslant \mathrm{E}(X^k) \leqslant U(k,n,p).$$
In particular I am interested in the case where $k$, $n$, $k/n$ are large and $p$, $kp$, $np$ are small. (Concrete values: $n\approx 10^3$, $k\approx 5\times 10^5$, $p\approx 10^{-16}$). Probably this translates into the limit $k\to\infty$, $kp\to 0$, $n\to \infty$, $k/n\to \infty$.
I have tried playing around with the recursion formula in Johnson, Kotz, Balakrishnan: Univariate Discrete Distributions (3rd ed. Wiley 2005), §3.3: $$\mathrm{E}(X^{k+1}) = \biggl[np + (1-p)p \frac{\mathrm{d}}{\mathrm{d}p}\biggr] \mathrm{E}(X^k),$$ but to no avail.
I have read this answer, based on Jensen's inequality, and it seems to suggest that $\mathrm{E}(X^k)\simeq n^k$ for $k$ large; in that case I'd need some extra terms in this asymptotic expansion. I have also read this answer, but I'm not sure that its series expansion is valid in my case.
Updated context: What I'm actually interested in is the ratio $\frac{\mathrm{E}(X^{k+1})}{\mathrm{E}(X^k)}$ in the asymptotic case described above. The first answer I mentioned seems to imply that this ratio is asymptotically equal to $n^{k+1}/n^k \equiv n$, but I'd like to find out the next correction to this approximation.
It turns out that the ratio $\frac{\mathrm{E}(X^{k+1})}{n\mathrm{E}(X^k)}$ is the posterior distribution of an exchangeable model for two events (e.g. coin tossing), with a binomial hyperparameter prior, conditioned on the observation of $k$ instances of the first event (say, Heads). When $k$ is large this posterior should give probability almost 1 to the first event, which is in accord with common sense and with the result above from Jensen's inequality. What interests me is the "almost", that is, the next correction to $\mathrm{E}(X^k)\simeq n^k$. I'm looking up more about the proof of Jensen's inequality, maybe I'll find something there.