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. 

Putting (probably far too many) details around Gerry Myerson's comment and avoiding the awkward upper bounds in IV_'s answer's sums...
\begin{align*}
f_n(x) &= \left( \sum_{k=0}^{n+1} x^{k} \right)^n  \\
    &= \left( \frac{1}{1-x} - \frac{x^{n+2}}{1-x} \right)^n  \\
    &= \sum_{k=0}^n (-1)^k \binom{n}{k} \left( \frac{1}{1-x} \right)^{n-k} \left( \frac{x^{n+2}}{1-x} \right)^k  \\
    &= \sum_{k=0}^n (-1)^k \binom{n}{k}  x^{(n+2)k} \left( 1-x \right)^{-n}  \\
    &= \sum_{k=0}^{n+1} (-1)^k \binom{n}{k} x^{(n+2)k} \sum_{d = 0}^{n(n+1) - (n+2)k}  \binom{n+d-1}{d} x^d  \\
    &= \sum_{i=0}^{n(n+1)} \left( \sum_{k=0}^{n+1} (-1)^k \binom{n}{k} \binom{n+i-(n+2)k-1}{i-(n+2)k} \right.\\
    &\qquad \left. \phantom{\sum_{i=0}^{n(n+1)}} \left[ 0 \leq i-(n+2)k \leq n(n+1)-(n+2)k \right] \right) x^i  \\
    &= \sum_{i=0}^{n(n+1)} \left( \sum_{k=0}^{n+1} (-1)^k \binom{n}{k} \binom{n+i-(n+2)k-1}{i-(n+2)k} \right.\\
    &\qquad \left. \phantom{\sum_{i=0}^{n(n+1)}} \left[ (n+2)k \leq i\right] \right) x^i  \text{.}
\end{align*}
Above, we have used (roughly in top-to-bottom order) geometric sums and series, the binomial theorem (twice), the fact that we need not retain powers of $x^d$ that end up producing powers above the largest power of $x$ in $f_n(x)$ (since such contributions will conspire to cancel to zero), forcing $d$ to satisfy $(n+2)k+d = i$, the Iverson bracket, and the facts that $(n+2)k \geq 0$ and $i$ is upper bounded.
There's really an exclusion-inclusion hiding here, fairly explicit in the fourth line.  As an example, per that line, 
$$ f_3(x) = \frac{1}{(1-x)^3} - \frac{3x^5}{(1-x)^3} + \frac{3x^{10}}{(1-x)^3} - \frac{x^{15}}{(1-x)^3}  \text{,}  $$
where we can see that what we want is the sum of four degree-shifted and scaled copies of polynomials constructed directly from the third diagonal of Pascal's triangle.  
We can also frame this example as counting lattice points in a finite cube.  We include all the points in triangular numbers until we reach the first corner (at $x^4$), where we need to subtract off contributions from the $3$ missing cubes starting at those corners, until they overlap -- causing double-deletion of points past the second "layer" of vertices -- and we add in $3$ cubes starting at those vertices, until we reach the last vertex and deal with the final batch of overlapping and overcounting.
These coefficients count lattice points in layers (of constant coordinate sum) of: 


*

*$n=1$: lattice points in layers of a line segment of length $3$, 

*$n=2$: lattice points in layers of a square of side $4$, 

*$n=3$: lattice points in layers of a cube of side $5$, 

*$n=4$: lattice points in layers of a tesseract of side $6$,

*$n=5$: lattice points in layers of a pentaract of side $7$,  

*and so on ...

