Prove that for all $k \in \mathbb N$ and $x \neq 0$ $h^{(k)}_k (x)=(-1)^k x^{-2k} h_k (x)$ 
For all $k \in \mathbb N$ let function $h_k: \mathbb  R\setminus  \left\{ 0\right\}  \rightarrow \mathbb R$ and $h_k(x)=x^{k-1}e^{\frac{1}{x}}$ for $x\neq 0$.Prove that for all $k \in \mathbb N$ and $x \neq 0$:  $$h^{(k)}_k(x)=(-1)^k x^{-2k} h_k(x)$$ where $h^{(k)}_k$ is $k^{th}$ derivative of $h_k$

My try:
Let: $$f(x)=x^{k-1}$$ $$g(x)=e^{\frac{1}{x}}$$ Then from General Leibniz Rule: $$h^{(k)}_k(x)=(f \cdot g)^{(k)} = \sum_{i=0}^k {k \choose i} f^{(i)} g^{(k-i)}$$However I have a problem because $g^{(k)}$ has really complicated formulas at further derivatives and it is too hard for me to get $h^{(k)}_k(x)=(-1)^k x^{-2k} h_k(x)$.
Is there any smart way to do this?
 A: Are you sure that there is nothing missing in your statement? Below is my work but I have an added term at the end... How does $h(x)$ have a dependence on $k$ if $k$ is marking the derivative term?
Using mathematical induction, let the induction statement be 
$$P(k): h^{(k)}(x) = (-1)^kx^{-2k}h(x)$$
The base case $P(1)$ states:
$$P(1) : h^{(1)}(x) = (-1)^1x^{-2(1)}h(x)$$
This is easily verified by taking the first derivative of $h(x)$.
$$h^{(1)}(x) = \dfrac{d}{dx}(x^{1-1}e^\frac{1}{x})$$
$$h^{(1)}(x) = \dfrac{d}{dx}(e^\frac{1}{x})$$
$$h^{(1)}(x) = -\dfrac{1}{x^2}e^\frac{1}{x}$$
$$h^{(1)}(x) = (-1)^1x^{-2(1)}x^{1-1}h(x)$$
$$h^{(1)}(x) = (-1)^1x^{-2(1)}h(x)$$
Now with the base case being demonstrated, assume $P(k)$ is true and show $P(k+1)$, which states
$$P(k+1): h^{(k+1)}(x) = (-1)^{k+1}x^{-2(k+1)}h(x)$$
Since we assume $P(k)$ to be true, we can say that
$$h^{(k+1)}(x) = \dfrac{d}{dx}h^k(x)$$
$$h^{(k+1)}(x) = \dfrac{d}{dx}((-1)^kx^{-2k}h(x))$$
$$h^{(k+1)}(x) = (-1)^k\dfrac{d}{dx}(x^{-2k}h(x))$$
$$h^{(k+1)}(x) = (-1)^k(-2kx^{-2k-1}h(x) + x^{-2k}h^{(1)}(x))$$
Substitution for $h^{(1)}(x)$ yields
$$h^{(k+1)}(x) = (-1)^k(-2kx^{-2k-1}h(x) + x^{-2k}(-1)^1x^{-2(1)}h(x))$$
$$h^{(k+1)}(x) = (-1)^{k+1}(2kx^{-2k-1}h(x) + x^{-2k}x^{-2(1)}h(x))$$
$$h^{(k+1)}(x) = (-1)^{k+1}(2kx^{-2k-1}h(x) + x^{-2k-2}h(x))$$
$$h^{(k+1)}(x) = (-1)^{k+1}x^{-2k-2}h(x)(2kx + 1)$$
$$h^{(k+1)}(x) = (-1)^{k+1}x^{-2(k+1)}h(x)(2kx + 1)$$
