Expanding this polynomial with a given power Is there possibly a more simple way to expand this?:
$(a+b+c+d+e+f+g+h+i)^{10}$.
Would you be able to use the binomial theorem? 
 A: There's a theorem called the multinomial theorem. It might be easier to write $x_1,x_2,\dots,x_9$ rather than $a,b,\dots,i$.
Then $$(x_1+\dots+x_9)^{10} = \sum_{(i_1,\dots,i_9)}_{i_1+\dots i_9=10} \binom{10}{i_1,\dots,i_9} x_1^{i_1}x_2^{i_2}\dots x_9^{i_9}$$
Where $$\binom{10}{i_1,\dots,i_9} = \frac{10!}{i_1!\dots i_9!}$$
So, the above sum is taken over all $9$-tuples of non-negative integers that add up to $10$.
You can probably prove the multinomial theorem using repeated applications of the binomial theorem, perhaps, but I wouldn't want to try.
A: No, typically the binomial theorem applies to binomials: "nomials" of the form $\;(x + y)^n\;$, with the "bi" in "binomial" meaning the sum of two arguments raised to a power. There is a very belabored means of using the binomial theorem by applying it repeatedly, but you might want to consider the following:
What you've got is a "multinomial".
See the Multinomial Theorem for some guidance. (It still involves a lot of work, though!)
From Wikipedia:

For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:
$$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}$$
where
$${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}$$
is a multinomial coefficient. The sum is taken over all combinations of nonnegative integer indices $k_1$ through $k_m$ such that the sum of all $k_i$ is $n$. That is, for each term in the expansion, the exponents of the $x_i$ must add up to n. Also, as with the binomial theorem, quantities of the form $x_0$ that appear are taken to equal 1 (even when $x$ equals zero).
In the case $m = 2,$ this statement reduces to that of the binomial theorem.

