Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1}$$ 
upon expanding them we get:
$$f_1(x)=1 + x + x^2$$
$$f_2(x)=1 + 2 x + 3 x^2 + 4 x^3 + 3 x^4 + 2 x^5 + x^6$$
$$f_3(x)=1 + 3 x + 6 x^2 + 10 x^3 + 15 x^4 + 18 x^5 + 19 x^6 + 18 x^7 + 
 15 x^8 + 10 x^9 + 6 x^{10} + 3 x^{11} + x^{12}$$
$$f_4(x)=1 + 4 x + 10 x^2 + 20 x^3 + 35 x^4 + 56 x^5 + 80 x^6 + 104 x^7 + 
 125 x^8 + 140 x^9 + 146 x^{10} + 140 x^{11} + 125 x^{12} + 104 x^{13} + 
 80 x^{14} + 56 x^{15} + 35 x^{16} + 20 x^{17} + 10 x^{18} + 4 x^{19} + x^{20}$$
$$\vdots$$
$$f_{n-1}(x)=1 + ?x + ?x^2 + ?x^3 +?x^4+ ?x^5+\cdots+?x^{n(n-1)}$$ 
I'm wondering how to determine the coefficients for the n-th order? I can observe that the coefficients are symmetric. 

We can think at the problem as a stars and bar problem, that uses the inclusion-exclusion principle, as in the answer cited by IV_ and this one.
We interprete
\begin{equation}(1+x+x^2 + \cdots + x^{n})^{n-1} = \underbrace{(1+x+x^2 + \cdots + x^{n})(1+x+x^2 + \cdots + x^{n})\cdots (1+x+x^2 + \cdots + x^{n})}_{n-1\text{ times }}\end{equation}
as follows. There are $n-1$ boxes, separated by $n-2$ bars. Since we are interested in the coefficient of $x^k$, we want to put $k$ indistiguishable balls in those $n-1$ boxes. You are allowed to take $n$ balls from each term of the product above, but no more.
Suppose you want to evaluate the coefficient of $x^{15}$ in
$$(1+x+x^2+x^3+x^4+x^5+x^6)^5,$$
without expanding the whole product. There are $5$ boxes separated by $5-1=4$ bars, in which we have to put $k=15$ balls, with the restriction that we may pick from one of the five terms in the product at most $6$ balls (the highest term is $x^6$). One example of that is
$$******\vert****\vert**\vert*\vert**,$$
which would be the contribution of $x^6$, $x^4$, $x^2$, $x$ and $x^2$ from the first, second, third, fourth and fifth term of the product. The number of ways to put $15$ indistinguishable balls in the $5$ boxes is
$${15 + 5-1\choose 5-1} = 3876.$$
Now you have to subtract the number of ways to put $15$ indistinguishable balls in $5$ boxes, so that in at least one box more than $6+1=7$ balls appear. You have ${5\choose 1}$ ways to choose the box in which to put the $7$ balls. The other $15-7 = 8$ balls can be placed in ${8 + 5-1\choose 5-1}$ ways. In total you have
$${5\choose 1}{8 + 5-1\choose 5-1} = 2475 $$
ways to put $15$ indistinguishable balls into $5$ boxes, so that in at least one box more than $7$ balls appear. 
Now, here comes the inclusion exclusion principle in play, you also did subtract the cases, where in more than $1$ container there were at least $7$ balls. So you add the number of ways to put $15$ indistinguishable balls into $5$ boxes, where in more than $2$ container there were at least $7$ balls. You have ${5\choose 2}$ ways to choose the containers where to put the $7$ balls and ${1 + 5 -1 \choose 5-1}$ ways to put the remaining ball in the other containers. So in total there are
$${5 \choose 2}{1+5-1\choose 5-1} = 50$$
ways to put the $15$ indistinguishable balls into $5$ boxes, where in more than $2$ container there were at least $7$ balls. 
The coefficient of $x^{15}$ in $(1+x+x^2+x^3+x^4+x^5+x^6)^5$ is therefore
$${15 + 5-1\choose 5-1} - {5\choose 1}{8 + 5 -1 \choose 5-1} + {5\choose 2}{1+5-1\choose 5-1} = 3876-2475+50 = 1451.$$
