Multiple Poisson distribution. When $n$ is large and $np_j = \lambda_j$ is moderate for $j=1, ... , r-1$, the multinomial distribution can be approximately by $$e^{-(\lambda_1+ \cdot\cdot\cdot + \lambda_{r-1})}\frac{\lambda_1^{k_1} \lambda_2^{k_2} \cdot\cdot\cdot \lambda_{r-1}^{k_{r-1}}}{k_1!k_2! \cdot\cdot\cdot k_{r-1}!}.$$
The multinomial distribution is $$\frac{n!}{k_1!k_2! \cdot\cdot\cdot k_{r-1}!}p_1^{k_1}\cdot\cdot\cdot p_{r-1}^{k_{r-1}}$$ where $k_1 + \cdot\cdot\cdot + k_{r-1} = n$, and $p_1 + \cdot\cdot\cdot +p_{r-1} =1$. Rearranging, I can see that $$\frac{\lambda_1^{k_1} \lambda_2^{k_2} \cdot\cdot\cdot \lambda_{r-1}^{k_{r-1}}}{k_1!k_2! \cdot\cdot\cdot k_{r-1}!} \frac{n!}{n^n}.$$
Therefore, the problem is reduced to showing $n!/n^n \to e^{\lambda_1 + \cdot\cdot\cdot \lambda_{r-1}}$, and I am stuck here. I guess it is something related to $a_{n+1}/ a_n \to e$ for $a_n = n^n/n!$, but I am not really sure.
I would appreciate if you give some help.
One issue is that it is unclear how the $k_1,\ldots, k_{r-1}$ vary with $n$. Since they sum to $n$, at least one of them must grow to infinity, so you cannot just focus on $n!/n^n$ alone. It is difficult to give a concrete statement about asymptotics without information about how they vary with $n$.
I believe the notation of "$r-1$" instead of $r$ may point to the intended interpretation where one $k_j$ tends to infinity while the others are fixed. For a multinomial with $r$ categories with respective probabilities $p_1, p_2, \ldots, p_{r-1}, 1-(p_1 + \cdots + p_r)$, the probability of having $k_1,k_2,\ldots, k_{r-1}, n-(k_1 + \cdots + k_{r-1})$ in each category is $$\frac{n!}{k_1! k_2! \cdots k_{r-1}! (n-(k_1 + \cdots + k_{r-1}))!} \frac{\lambda_1^{k_1} \lambda_2^{k_2} \cdots \lambda_{r-1}^{k_{r-1}} (n-(\lambda_1 + \cdots + \lambda_{r-1}))^{n-(k_1 + \cdots + k_{r-1})}}{n^n},$$ where $\lambda_i = np_i$.
Stirling's approximation gives $$\frac{n!}{(n-(k_1 + \cdots + k_{r-1}))!} \sim \sqrt{\frac{n}{n-(k_1 + \cdots + k_{r-1})}} \frac{n^n}{[n-(k_1 + \cdots + k_{r-1})]^{n-(k_1 + \cdots + k_{r-1})}} e^{-(k_1 + \cdots + k_{r-1})}$$
so the probability is asymptotically equivalent to
$$\frac{\lambda_1^{k_1} \lambda_2^{k_2} \cdots \lambda_{r-1}^{k_{r-1}} (n-(\lambda_1 + \cdots + \lambda_{r-1}))^{n-(k_1 + \cdots + k_{r-1})}}{k_1! \cdots k_{r-1}! [n-(k_1 + \cdots + k_{r-1})]^{n-(k_1 + \cdots + k_{r-1})}} \sqrt{\frac{n}{n-(k_1 + \cdots + k_{r-1})}} e^{-(k_1 + \cdots + k_{r-1})}.$$
Taking the limits $\sqrt{\frac{n}{n-(k_1 + \cdots + k_{r-1})}} \to 1$ and $$\frac{[n-(\lambda_1 + \cdots + \lambda_{r-1})]^{n-(k_1 + \cdots + k_{r-1})}}{[n-(k_1 + \cdots + k_{r-1})]^{n-(k_1 + \cdots + k_{r-1})}} = \frac{[1-(\lambda_1 + \cdots + \lambda_{r-1})/n]^{n-(k_1 + \cdots + k_{r-1})}}{[1-(k_1 + \cdots + k_{r-1})/n]^{n-(k_1 + \cdots + k_{r-1})}} \sim \frac{[1-(\lambda_1 + \cdots + \lambda_{r-1})/n]^{n}}{[1-(k_1 + \cdots + k_{r-1})/n]^{n}} \to \frac{e^{-(\lambda_1 + \cdots + \lambda_{r-1})}}{e^{-(k_1+ \cdots + k_{r-1})}}$$ leads to $$\frac{\lambda_1^{k_1} \lambda_2^{k_2} \cdots \lambda_{r-1}^{k_{r-1}} }{k_1! \cdots k_{r-1}! } e^{-(\lambda_1 + \cdots + \lambda_{r-1})}.$$