Let \begin{align} c_k:=\sum_{n=1}^k \sum_{k=k_1+\dotsb+k_n}\frac{1}{n!}\frac{1}{k_1\cdots k_n} \end{align} where the second sum is over positive integers $k_j$. I need to prove that $\sqrt[k]{c_k}\to1$ as $k\to\infty$.
To simplify the problem one might consider the obvious bound \begin{align} \sum_{n=1}^k \sum_{k=k_1+\dotsb+k_n}\frac{1}{n!}=\sum_{n=1}^k \frac{1}{n!}\binom{k-1}{n-1}. \end{align}
You may consider the generating function: $C(t) = \sum_{k\geq 1} c_k t^k$, and write $t^k = t^{k_1} \cdots t^{k_n}$ to carry out the interior sums. The end result is (surprisingly) simple: $C(t)=t/(1-t)$.
But I think it would spoil the fun if I wrote down all the intermediate calculations. Better try yourself and ask if encountering problems.
Edit: You should insert details in the following calculation:
$$ 1+ C(t) = 1+ \sum_{n\geq 1} \frac{1}{n!} \left( \sum_{k_1\geq 1} \frac{t^{k_1}}{k_1} \cdots \sum_{k_n\geq 1} \frac{t^{k_n}}{k_n}\right) = \sum_{n\geq 0} \frac{1}{n!} \left( \ln \frac{1}{1-t} \right)^n = \exp\left(\ln \left( \frac{1}{1-t} \right)\right)= \frac{1}{1-t}$$
