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  use  the  coefficient of operator  $[x^k]$  to denote the  coefficient of $x^k$  of  a series.

We obtain for $0\leq k\leq n(n-1)$:
  \begin{align*}
\color{blue}{[x^{k}]}&\color{blue}{\left(1+x+x^2+\cdots+x^n\right)^{n-1}}\\
&=[x^k]\left(\frac{1-x^{n+1}}{1-x}\right)^{n-1}\tag{1}\\
&=[x^k]\sum_{j=0}^\infty\binom{-(n-1)}{j}(-x)^j\sum_{l=0}^{n-1}\binom{n-1}{l}\left(-x^{n+1}\right)^l\tag{2}\\
&=[x^k]\sum_{j=0}^\infty\binom{n-j-2}{j}x^j\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{3}\\
&=\sum_{j=0}^{k}\binom{n-j-2}{j}[x^{k-j}]\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{4}\\
&=\sum_{j=0}^{k}\binom{n-k+j-2}{k-j}[x^{j}]\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{5}\\
&=\sum_{j=0}^{\left\lfloor\frac{k}{n+1}\right\rfloor}\binom{n-k+(n+1)j-2}{k-(n+1)j}[x^{(n+1)j}]\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{6}\\
&\,\,\color{blue}{=\sum_{j=0}^{\left\lfloor\frac{k}{n+1}\right\rfloor}\binom{(n+2)j-k-2}{k-(n+1)j}\binom{n-1}{j}(-1)^j}\tag{7}\\
\end{align*}

Comment:


*

*In (1) we use the finite  geometric series formula.

*In  (2)  we use the  binomial  series expansion and apply the binomial theorem.

*In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (4) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and set  the upper  limit of the outer sum to $k$ since indices $j>k$ do not contribute.

*In (5) we reverse the order of summation of the outer sum: $j\to k-j$.

*In (6) we consider only $(n+1)$-multiples of $j$ since other values do not occur as exponent of $x$ in the inner sum.

*In (7) we finally select the coefficients of $x^{(n+1)j}$ by taking $l=j$.
