Let $k_n:=\kappa_n$, $a_n:=E(X-EX)^n$, $b_n:=E|X-EX|^n$, so that $|a_n|\le b_n$. We have to show that
$$|k_n|\le n^n b_n$$
for natural $n$. For $n=1,2$ this is obvious. The key is the recursion
$$k_n=a_n-\sum_{m=1}^{n-1}\binom{n-1}{m-1}k_m a_{n-m}$$
at the end of this section , which implies
$$|k_n|\le b_n+\sum_{m=1}^{n-1}\binom{n-1}{m-1}|k_m| b_{n-m}.$$
This allows us to use the induction on $n$, which implies
$$|k_n|\le l_n b_n,$$
where
$$l_n:=1+\sum_{m=1}^{n-1}\binom{n-1}{m-1}m^m;$$
here we used the log-convexity of $b_j$ in $j$ together with the fact $b_0=1$ (or, equivalently, Hölder's inequality), which implies $b_m b_{n-m}\le b_n$.

It remains to show that $l_n\le n^n$ for $n\ge2$. This is trivial for $n=2$. For $n\ge3$, proceed again by induction:
\begin{align*}
l_n&=1+\sum_{m=1}^{n-1}\binom{n-2}{m-1}m^m+\sum_{m=1}^{n-1}\binom{n-2}{m-2}m^m \\
&=1+\sum_{m=1}^{n-2}\binom{n-2}{m-1}m^m+(n-1)^{n-1}+\sum_{m=2}^{n-1}\binom{n-2}{m-2}m^m \\
&=l_{n-1}+(n-1)^{n-1}+\sum_{j=1}^{n-2}\binom{n-2}{j-1}(j+1)^{j+1} \\
&\le l_{n-1}+(n-1)^{n-1}+\sum_{j=1}^{n-2}\binom{n-2}{j-1}j^j c_n \\
&=l_{n-1}+(n-1)^{n-1}+(l_{n-1}-1) c_n \\
&\le (n-1)^{n-1}+(n-1)^{n-1}+(n-1)^{n-1}c_n \\
&=(n-1)^{n-1}\Big(2+\frac{(n-1)^{n-1}}{(n-2)^{n-2}}\Big)=:r_n
\end{align*}
where
$$c_n:=\max_{1\le j\le n-2}\frac{(j+1)^{j+1}}{j^j}=\frac{(n-1)^{n-1}}{(n-2)^{n-2}},$$
because
$$\frac{(j+1)^{j+1}}{j^j}=\Big(1+\frac1j\Big)^j(j+1)$$
is increasing in $j\ge1$.

It remains to check that $r_n\le n^n$ for all real $n>2$, which, by substitution $n=1+\frac1t$, can be rewritten as
\begin{equation}
f_1(t)f_2(t)\le1, \tag{1}
\end{equation}
where
\begin{equation*}
f_1(t):=e \left(\frac{1}{t+1}\right)^{\frac{1}{t}+1},\quad
f_2(t):=\frac{\left(\left(\frac{1}{t}\right)^{\frac{1}{t}}
\left(\frac{1}{t}-1\right)^{\frac{t-1}{t}}+2\right) t}{e};
\end{equation*}
everywhere here, $t\in(0,1]$.

To prove (1), it is enough to prove
\begin{equation}
f_1\le g_1 \tag{2},
\end{equation}
\begin{equation}
f_2\le g_2 \tag{3},
\end{equation}
\begin{equation}
g_1g_2\le1 \tag{4}
\end{equation}
on $(0,1]$, where
\begin{equation*}
g_1(t):=\frac{7 t^2}{24}-\frac{t}{2}+1,\quad
g_2(t):=\left(\frac{2}{e}-\frac{1}{2}\right) t+1.
\end{equation*}

To prove (2), consider
\begin{equation*}
d_1:=\ln f_1-\ln g_1
\end{equation*}
and then $d_{11}(t):=t^2d_1'(t)$ and $d_{11}'$, which latter is a rational function, which is easily seen to be $<0$. So, $d_{11}$ decreases. Also, $d_{11}(0+)=0$. So, $d_{11}<0$ (on $(0,1]$) and hence $d_1$ decreases. Also, $d_1(0+)=0$. So, $d_1<0$, which yields (2).

To prove (3), rewrite it as

\begin{equation*}
d_2(t):=\ln \left(\left(\frac{1}{t}-1\right)^{\frac{t-1}{t}}
\left(\frac{1}{t}\right)^{\frac{1}{t}}\right)-\ln \left(\frac{e (2-t)}{2 t}\right)\le0
\end{equation*}
(for $t\in(0,1]$).
Consider then $d_{21}(t):=t^2d_2'(t)$ and $d_{21}'$, which latter is a rational function, which is easily seen to be $<0$ on $(0,1)$. So, $d_{21}$ decreases. Also, $d_{21}(0+)=0$. So, $d_{21}<0$ (on $(0,1]$) and hence $d_2$ decreases. Also, $d_2(0+)=0$. So, $d_2<0$, which yields (3).

Finally, (4) is elementary, since $g_1g_2$ is a polynomial (of degree $3$). $\Box$