Order of operations in polynomial with exponent I have a simple question about whether or not my approach is correct in simplifying a polynomial, here it is,
$(n(n+1)/2)^2 = ((n^2+n)/2)^2 = 1/4(n^4+2n^3+n^2)$
I apologize if you find that hard to read, I can't write the equations the way I would like.
Many thanks.
 A: It is correct, but if desired to write with no parenthesis you may opt to distribute the $\frac{1}{4}$ through the expression to arrive at $$\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}$$
A small piece of advice for if you are faced with exponents of higher power than $2$.
There is such a thing known as the Binomial Theorem.
It says:

$(x+y)^n = x^n+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+\dots+\binom{n}{r}x^{n-r}y^r+\dots+\binom{n}{n-1}x^1y^{n-1}+y^n$

or more concisely as:

$(x+y)^n = \sum\limits_{r=0}^n \binom{n}{r}x^{n-r}y^r$

If you don't know easily what the symbols $\binom{n}{r}$ mean, that's okay for now.  You can figure the small ones out by making Pascals triangle:

To create the next row, add the two numbers that appear above it (except for the far left or far right in which case there is only a one above it and empty space).  For example, in the fifth line, the $10$ was created by adding the $4$ and the $6$ above it.
These numbers are the coefficients of having expanded $(x+y)^n$
$$\begin{array}{l&c} (x+y)^0=&1\\
(x+y)^1=&1x+1y\\
(x+y)^2=&1x^2+2xy+1y^2\\
(x+y)^3=&1x^3+3x^2y+3xy^2+1y^3\\
(x+y)^4=&1x^4+4x^3y+6x^2y^2+4xy^3+1y^4\\
(x+y)^5=&1x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+1y^5\\
\vdots\end{array}$$
In your case, since the two pieces are $(\color{red}{n^2}+\color{blue}{n})$, you have $(n^2+n)^2 = (\color{red}{n^2})^2 + 2\color{red}{n^2}\color{blue}{n}+\color{blue}{n}^2$
If you were faced with something like $(n^2+n)^4$ it would be 
$$(n^2+n)^4=(\color{red}{n^2})^4+4(\color{red}{n^2})^3\color{blue}{n}+6(\color{red}{n^2})^2\color{blue}{n}^2+4\color{red}{n^2}\color{blue}{n}^3+\color{blue}{n}^4\\=n^8+4n^7+6n^6+4n^5+n^4$$
