Taylor series proof by induction I am kind of stuck with this problem. I know the basics of induction proof and how to use it to prove stuff about basic arithmetic and geometric series and so on, but the problem below seems a bit new and unfamiliar to me.
The problem is about proving how a Taylor series develops using induction proof.
The problem goes like this:
Given the function $f(x)=\frac{1}{x^2}$
Prove by induction that $f^{(k)}(x)=\frac{(-1)^{k}(k+1)!}{x^{k+2}}$ for all $k\geq 0$.
Is there anyone out there who is willing to provide a solution to this problem and help me and expand my horizon/ help in my learning?
All help is welcomed and appreciated.
 A: Let $g(k)=f^{(k)}(x)=\frac{(-1)^k(k+1)!}{x^{k+2}}$
For $g(0)$, $f(x)=\frac{1×1!}{x^2}$
$g(1)=\frac{\partial(x^{-2})}{\partial{x}}=\frac{-2}{x^3}=\frac{(-1)(2)!}{x^{3}}$
Therefore $g(k)$ is true for $k=0$ and $1$.
Let $g(k)=f^{(k)}(x)=\frac{(-1)^k(k+1)!}{x^{k+2}}$ be true for all $k≥2$.
Therefore,$$g(p)=f^{(p)}(x)=\frac{(-1)^p(p+1)!}{x^{p+2}}$$
And
$$g(p+1)=f^{(p+1)}(x)=\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}$$$\Rightarrow\frac{\partial g(p)}{\partial x}=\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}\Rightarrow\frac{\partial{\frac{(-1)^{p}(p+1)!}{x^{p+2}}}}{\partial x}=\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}$
Now we need to calculate $$\frac{\partial{\frac{(-1)^{p}(p+1)!}{x^{p+2}}}}{\partial x}$$
Taking the constants out, we are left with $$(-1)^p(1+p)!\frac{\partial{(x^{-p-2}})}{\partial x}$$
This gives us $$(-1)^p(1+p)!(-2-p)(x^{-p-3})$$$\Rightarrow $$(-1)^p(1+p)!(-1)(p+2)\frac{1}{x^{p+3}}\Rightarrow\frac{(-1)^{p+1}(p+2)!}{x^{p+3}}$
Therefore for every $p$, $g(p)$ as well as $g(p+1)$ is true.
Hence proved.
I am not very sure about what I have done is correct or not.... Maybe there is some mistake. For that pardon me.
