Find an asymptotically tight estimate for the sum
$$
A\_n^{k}(\lambda)= \sum\_{
\substack{a\_i\geq \lambda\_i
\\
a\_1+a\_2+\dots a\_k=n
}} \prod\_{i=1}^k a\_i!
$$


Is the leading term going to be
$$|\textrm{Number of Maximal Lambda}|(j-\lambda\_1-\lambda\_2-\lambda\_3- \lambda\_4+ \lambda\_{max})!\frac{1}{\lambda\_{\max}!}\prod\_{i=1}^4 \lambda\_i!
$$


Edit: As of right now there is some discrepency as to weather this conjecture is correct or not. 


This question has been asked before in the binomial situation [Here](https://math.stackexchange.com/questions/151441/calculate-sums-of-inverses-of-binomial-coefficients)



Sum of inverse of multinomial coefficients
I add two hopefully useful remarks.


I consider the general situation. In the sequel $k\geq 1$ and $\lambda=(\lambda\_1,\ldots,\lambda\_{k+1})$ are fixed, $s:=\sum\_{i=1}^{k+1}\lambda\_i$ and $n\geq s$.


Let $A\_n^{(k+1)}(\lambda):=\sum\limits\_{\stackrel{a\_i\geq \lambda\_i, i=1,\ldots,k+1}{a\_1+\ldots +a\_{k+1}=n}} \prod\_{i=1}^{k+1} a\_i!$ 
and let $\mathcal{S}\_{k}:=\{ (x\_1,\ldots,x\_{k+1})\;:\;x\_i\geq 0, \sum\_{i=1}^{k+1} x\_i=1\}$ denote the $k$-dimensional standard simplex.


(i) $A\_n^{(k+1)}(\mathbf{\lambda})$ has a nice geometric representation:


$$ A\_n^{(k+1)}(\lambda)=\frac{(n-s+k)!\,(n+k)!}{(n-s)!} \int\_{\mathcal{S}\_{k}} \int\_{\mathcal{S}\_{k}} \left(x\_1y\_1+\ldots+x\_{k+1}y\_{k+1}\right)^{n-s}\,\prod\_{i=1}^{k+1} x\_i^{\lambda\_i}d\mathbf{x}\,\,d\mathbf{y} $$


Proof: recall
 Dirichlet's integral: $$((\sum\_{i=1}^{k+1}a\_i)+k)!\, \int\_{\mathcal{S}\_{k}} \prod\_{i=1}^{k+1} x\_i^{a\_i}\,d\mathbf{x}=\prod\_{i=1}^{k+1} a\_i!$$
Now expand the integrand using the multinomial theorem and use Dirichlet's integral repeatedly.


EDIT: I have replaced the previous (unnecessarily complicated) derivation and again corrected the factor (apologies). (The conclusions remain unchanged).


(ii) Thus the large $n$ asymptotic of $A\_n^{(k+1)}(\lambda)$ can be investigated along the lines of the Laplace method (explained e.g. in chapter 4 of de Bruijn's book).


Remark: I have done that, and (unfortunately?) the result confirmed your original conjecture. Thus either my analysis or the accepted answer (or both) are incorrect (no offence). 


EDIT:


The Laplace analysis follows the usual scheme:
 (1) identify maxima
(2) cutoff tails, approximate integrand (3) complete tails  

I sketch the main steps. Let $I$ denote the integral above.


Basic intuition:


(1) Consider the scalar product $\langle \mathbf{x},\mathbf{y}\rangle>=\sum\_{i=1}^{k+1}x\_iy\_i\leq 1$. On the domain of integration $\frac{1}{k+1}\leq \langle\mathbf{x},\mathbf{y}\rangle\leq 1$, and $\langle\mathbf{x},\mathbf{y}\rangle$ can be 1 only in a "corner" $x\_iy\_i=1$ for a certain $i$.


(2) For any $0<\epsilon<1$ the part of $I$ where $\langle\mathbf{x},\mathbf{y}\rangle<1-\epsilon$ is asymptotically negligible (compared to its complement). Asymptotically thus only the parts around the corners (i.e. $x\_iy\_i\approx 1$) need to be considered.


With this in mind:
(3) choose $\epsilon$ so small that the corners $C\_i:=\{ x\in \mathcal{S}\_k\;:\; x\_i\geq 1-\epsilon, y\_i\geq 1-\epsilon\}$ are disjoint, and discuss their influence individually.


E.g. for corner $k+1$ choose $x\_1,\ldots,x\_k$ resp. $y\_1,\ldots,y\_k$ as independent coordinates. 
Then $x\_{k+1}=1-\sum\_{i=1}^k x\_i=:1-u$, $y\_{k+1}=1-\sum\_{i=1}^k y\_i=:1-t$.


We have $\langle\mathbf{x},\mathbf{y}\rangle=(1-u)(1-t)+\sum\_{i=1}x\_iy\_i\leq (1-u)(1-t)+ut$, thus $(\langle\mathbf{x},\mathbf{y}\rangle)^2\leq (1-2u(1-u))(1-2t(1-t))$. 
The parts of the integral where $u\geq 1/n^{2/3}$ or $t\geq 1/n^{2/3}$
are negligible because $\langle\mathbf{x},\mathbf{y}\rangle^n=\mathcal{O}(\exp(-n^{1/3}))$ there.
On the remaining part $\langle\mathbf{x},\mathbf{y}\rangle^n =\exp(-n(u+t))(1+o(1))$ uniformly.
Therefore introduce new coordinates $x\_1,\ldots,x\_{k-1},u$, $y\_1,\ldots,y\_{k-1},t$, then
\begin{align\*}
&\int\_{{C}\_{k+1}}\int \left(x\_1y\_1+\ldots+x\_{k+1}y\_{k+1}\right)^n\,\prod\_{i=1}^{k+1} x\_i^{\lambda\_i}d\mathbf{x}\,\,d\mathbf{y} \\
&=\int\limits\_{{x\_i\geq 0, x\_1+\ldots +x\_k=u}\atop{{y\_i\geq 0, y\_1+\ldots +y\_k=t}\atop {0\leq u,t\leq n^{-2/3}}}} e^{-n(u+t)}\prod\_{i=1}^k x\_i^{\lambda\_i}(1-u)^{\lambda\_{k+1}} dx\_1\ldots,dx\_k dy\_1\ldots dy\_k\,du\,dt\big(1+o(1)\big)\\
&= \int\_0^{n^{-2/3}}\int\_0^{n^{-2/3}} e^{-n(u+t)} (\frac{u}{n})^{s-\lambda\_{k+1}+k-1} \frac{\prod\_{i=1}^k\lambda\_i!}{(s-\lambda\_{k+1}+k-1)!} (1-u)^{\lambda\_{k+1}} 
(\frac{t}{n})^{k-1}\frac{1}{(k-1)!}\,du\,dt\big(1+o(1)\big)\\
&=\frac{\prod\_{i=1}^k \lambda\_i!}{n^{2k+s-\lambda\_{k+1}}}\left(1+o(1)\right)
\end{align\*}
where last line follows after substituting $z=nu,w=nt$ and completing the tails.
Summing over all corners one now gets your formula (after applying Stirling's formula to it, and up to the additional factors).


