Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.

The hard way, by Taylor:
First establish the derivatives,
$$\begin{align}
f(x)(x-1)&=x^6-1\\
f'(x)(x-1)+f(x)&=6x^5\\
f''(x)(x-1)+2f'(x)&=30x^4\\
f'''(x)(x-1)+3f''(x)&=120x^3\\
f''''(x)(x-1)+4f'''(x)&=360x^2\\
f'''''(x)(x-1)+5f''''(x)&=720x\\
f''''''(x)(x-1)+6f'''''(x)&=720\\
f'''''''(x)(x-1)+7f''''''(x)&=0\\
\dots\\
f^{(n+1)}(x)(x-1)+(n+1)f^{(n)}(x)&=0\\
\end{align}$$
Then solve for $x=0$:
$$\begin{align}
f(0)&=1,\\
f'(0)=f(0)&=1,\\
f''(0)=2f'(0)&=2,\\
f'''(0)=3f''(0)&=3!\\
f''''(0)=4f'''(0)&=4!\\
f'''''(0)=5f''''(0)&=5!,\\
f''''''(0)=6f'''''(0)-720&=0\\
\dots\\f^{(n)}&=0\end{align}$$
Conclusion,
$$f(x)=1+x+x^2+x^3+x^4+x^5.$$
