Originally Posted by

**adkinsjr** I know, but I don't see why that's true if $\displaystyle (n+1)!=n!(n+1)$ I obtain:

$\displaystyle \frac{2\cdot 3 \cdot 4\cdot\cdot\cdot (n+2)x^{n+1}}{n!(n+1)}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)x^n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}\frac{(n+2)x}{(n+1)(n+1)}=\frac{(n+2)x}{(n+1)^2 }$

So I cancel out the $\displaystyle 2\cdot 3\cdot 4\cdot\cdot\cdot x^nn!$

Notice that there are TWO $\displaystyle (n+1)$ factors in the denominator. Where does that second factor go?