How to prove that ${l \choose a_1,...,a_n}\le n^{l-1} $ , when $a_1+...+a_n=l$. In the proof of (Corollary 8 chap. 3 ) in the book "Sobolev Spaces on Domains" by Burenkov the following inequality is used : given $a_1,...,a_n \in \mathbb{N}$ such that $a_1+...+a_n=l$, then
$${l \choose a_1,...,a_n}\le n^{l-1}  $$
where 
$${l \choose a_1,...,a_n}=\frac{l!}{a_1!a_2!\cdots a_n!}$$
is the multinomial coefficient. How can one prove this fact? I was able to prove using multinomial theorem that
$${l \choose a_1,...,a_n}\le n^{l}$$
but I couldn't prove the sharper inequality.  
 A: Assuming that $a_1,\ldots,a_n$ are distinct integers,
$$n\binom{l}{a_1,\ldots,a_n}\\=\binom{l}{a_1,a_2,\ldots,a_{n-1},a_n}+\binom{l}{a_2,a_3\ldots,a_{n},a_1}+\ldots+\binom{l}{a_n,a_1,\ldots,a_{n-2},a_{n-1}}$$
but the last sum is less than the sum of any multinomial coefficient $\binom{l}{x_1,x_2,\ldots,x_{n-1},x_n}$ with $x_1+x_2+\ldots+x_{n-1}+x_n=n$, hence it is less than $n^l$. The same argument gives the stronger:
$$\binom{l}{a_1,\ldots,a_n} \leq \frac{n^l}{n!}.$$
We may deal with the cases in which $a_i=a_j$ by inclusion-exclusion.
A: The inequality $$\binom{l}{a_1,a_2,\ldots,a_n}\leq n^{l-1}$$
holds for positive integers $n$ and $l$ and for all integers $a_1,a_2,\ldots,a_n$ such that $a_1+a_2+\ldots+a_n=l$ (where we interpret $\dbinom{l}{a_1,\ldots,a_n}$ as $0$ if any of the $a_i$'s is negative).  The equality holds if and only if
(1) $n=1$,
(2) $n>1$, $l=1$, and $\left(a_1,a_2,\ldots,a_n\right)$ is a permutation of $(1,\underbrace{0,\ldots,0}_{(n-1)\,\text{zeros}})$, or 
(3) $n=2$, $l=2$, and $\left(a_1,a_2\right)=(1,1)$.
Note that $$\binom{l}{a_1,a_2,\ldots,a_n}=\binom{l-1}{a_1-1,a_2,\ldots,a_n}+\binom{l-1}{a_1,a_2-1,\ldots,a_n}+\ldots+\binom{l-1}{a_1,a_2,\ldots,a_n-1}\,.$$
We can then prove the inequality by induction on $l$, with the trivial base case $l=1$.
