The question that follows is a continuation of this previous Stage $1$ question needed as part of a derivation of the Associated Legendre Functions Normalization Formula: $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}$$ where for each $m$, the functions $${P_L}^m(x)=\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\tag{1}$$ are a set of Associated Legendre functions on $[−1, 1]$.
The question in my textbook asks me to
Write $(1)$ with $m$ replaced with $-m$ then use the fact that $$\begin{align}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L&=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\end{align}\quad \longleftarrow\text{(Stage 1)}$$ to show that $${P_L}^{-m}(x)=(-1)^m\frac{(L-m)!}{(L+m)!}{P_L}^{m}(x)\tag{2}$$
Start of attempt:
$$\begin{align}\require{enclose}{P_L}^{-m}(x)&=
\frac{1}{2^LL!}\left(1-x^2\right)^{-m/2}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\\&=
\frac{1}{2^LL!}\left(1-x^2\right)^{-m/2}\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\quad\quad\longleftarrow\bbox[#AFA]{\text{Using Stage 1}}\\&=
\frac{(-1)^{-m/2}}{2^LL!}\left(x^2-1\right)^{-m/2}\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\\&=
\frac{(-1)^{-m/2}}{2^LL!}\left(x^2-1\right)^{m/2}\frac{(L-m)!}{(L+m)!}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\\&=
\frac{\color{#F8F}{\enclose{updiagonalstrike}{\color{black}{(-1)^{-m/2}}}}\cdot\color{#F8F}{\enclose{updiagonalstrike}{\color{black}{(-1)^{m/2}}}}}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\frac{(L-m)!}{(L+m)!}\\&=
\frac{(L-m)!}{(L+m)!}\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\\&=
\frac{(L-m)!}{(L+m)!}{P_{L}}^m(x)\quad\quad\longleftarrow\bbox[#FFA]{\text{By substituting Equation (1)}}\\&\ne
(-1)^m\frac{(L-m)!}{(L+m)!}{P_L}^{m}(x)\end{align}$$
End of attempt.
So basically I am missing a factor of $(-1)^m$ as the $(-1)^{-m/2}$ and $(-1)^{m/2}$ cancelled each other out, and I'm therefore unable to prove Equation $(2)$. I have checked the proof for errors and am unable to find any. I am guessing that I have overlooked something simple thus making a trivial mistake somewhere along the line.
Is anyone able to locate and explain where I have made the error?
EDIT:
Taking the advice of the answer given by Markus:
$$\begin{align}\require{enclose}{P_L}^{-m}(x)&= \frac{1}{2^LL!}\left(1-x^2\right)^{-m/2}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\\&= \frac{1}{2^LL!}\left(1-x^2\right)^{-m/2}\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\quad\quad\longleftarrow\bbox[#FA8]{\text{Using Stage 1}}\\&= \frac{1}{2^LL!}(-1)^m\left(1-x^2\right)^m\left(1-x^2\right)^{-m/2}\frac{(L-m)!}{(L+m)!}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\\&= (-1)^m\frac{(L-m)!}{(L+m)!}\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\\&= (-1)^m\frac{(L-m)!}{(L+m)!}{P_{L}}^m(x)\quad\quad\longleftarrow\bbox[#AFF]{\text{By substituting Equation (1)}}\end{align}$$ as required ..... but only thanks to Markus :) ..... and not me :(
In this case the rule valid for real $a,b\geq 0$, namely \begin{align*} \sqrt{ab}=\sqrt{a}\sqrt{b} \end{align*} does no longer hold.
See for instance this question which lies at the heart of the problem.