Proof multinomial coefficient is always no less than 1 by induction The goal is to show 
$$\binom{n}{k}:=\frac{n!}{k_1!k_2!...k_d!}\geq1$$
where $k=(k_1,k_2,..,k_d)$ multiindex of dimension $d$ with $|k|=n$.
It is easy to show the case for $d = 1,2$.
$d = 1 $ is trivial.
For $d = 2$, let's use the common symbols
$$\frac{n!}{k!(n-k)!} = \frac{n(n-1)\dots(n-k+1)}{k!}\geq 1$$
since $(n-k)!$ cancel out and $n\geq k$.
The last step would be assume the statement valid for the case $d=p$.
Then for the case $d=p+1$, is it safe to claim that for each $k=(k_1,k_2,..,k_{p+1})$, there exist a $\tilde{k}=(\tilde{k}_1,..,\tilde{k}_p)$ such that
$$\tilde{k}_i = k_i\, , i=1,..,p-1$$
and 
$$\tilde{k}_p = k_p+k_{p+1}.$$
Then from(actually the binomial case)
$$\frac{\tilde{k}_p}{k_p!k_{p+1}!}\geq1$$
the proof is complete.
Are there any flaws hidden in the proof?
Thanks.
 A: You proof seems OK. But I would write it as
$$
\binom n{k_1,\ldots,k_d}
=\binom {k_1+k_2\cdots+k_d}{k_1}\binom{k_2+\cdots+k_d}{k_2}\ldots\binom{k_{d-1}+k_d}{k_{d-1}},
$$
a product of binomial coefficients all${}\geq1$, so the result is integer and${}\geq1$. The identity I wrote follows by induction from
$$
  \binom {k_1+k_2\cdots+k_d}{k_1,\ldots,k_d}
=\binom {k_1+k_2\cdots+k_d}{k_1,k_2,\cdots,k_{d-2},(k_{d-1}+k_d)}\binom{k_{d-1}+k_d}{k_{d-1}}.
$$
The multinomial coefficient after the "=" in this last equation is what your proof applies induction to, renaming the lower indices to $\tilde k_1,\ldots,\tilde k_{d-1}$ (only the final one has changed from the $k_i$).
The advantage of the formulation I gave is mainly that it shows you have found out a bit more about multinomial coeffcients than just their positivity (their combinatorial interpretation shows that they are integer).
A: This is a nice proof sketch.  It doesn't have flaws so much as missing details.  How are you combining the inductive hypothesis with the binomial case result?
