Deriving the Normalization formula for Associated Legendre functions: Stage $4$ of $4$ The question that follows is the final stage of the previous $3$ stages found here: Stage 1, Stage 2 and Stage 3 which are needed as part of a derivation of the Associated Legendre Functions Normalization Formula: $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}\tag{1}$$ where for each $m$, the functions $${P_L}^m(x)=\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\tag{2}$$ are a set of Associated Legendre functions on $[−1, 1]$.
The question in my textbook asks me to

Derive $(1)$ as follows: Multiply together the two formulas for ${P_{L}}^m(x)$ given by
  $(2)$ and $${P_L}^{m}(x)=(-1)^m\frac{(L+m)!}{(L-m)!}\frac{1}{2^LL!}\left(1-x^2\right)^{-m/2}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\quad\longleftarrow\text{(Stage 3)}$$ Then integrate by parts repeatedly lowering the $L+m$ derivative and raising the $L−m$ derivative until both are $L$ derivatives. Then use the regular Normalization formula for Legendre Functions: $$\displaystyle\int_{x=-1}^{1}[{P_{L}}(x)]^2\,\mathrm{d}x=\frac{2}{2L+1}\tag{3}$$ where ${P_{L}}(x)$ represents a Legendre function and ${P_{L}}^m(x)$ represents an associated Legendre function.


Start of attempt:
Multiplying together $(2)$ and the Stage $3$ formula yields:
$$[{P_{L}}^m(x)]^2=\frac{(-1)^m}{(2^LL!)^2}\frac{(L+m)!}{(L-m)!}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\tag{4}$$ Multiplying both sides of $(4)$ by $\mathrm{d}x$ and integrating gives:
$$\int[{P_{L}}^m(x)]^2\,\mathrm{d}x=\frac{(-1)^m}{(2^LL!)^2}\frac{(L+m)!}{(L-m)!}\color{red}{\int\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\,\mathrm{d}x}\tag{5}$$
Focusing now on the part marked $\color{red}{\mathrm{red}}$ and integrating by parts:
$$\int\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\,\mathrm{d}x$$
$$=\left.\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\right|_{-1}^1-\int\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\frac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x$$
$$=0-\int\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\frac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x$$
End of attempt.

I don't know how to take this calculation any further as I have no idea how to evaluate
$$\color{#180}{\int\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\frac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x}$$ 
Could someone please help me reach equation $(1)$ and finally end this derivation of 
$$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}\tag{1}$$

EDIT:
The Latex didn't render correctly in the description below the bounty; so I type it here instead:
One user has already given a detailed answer to this question that uses mathematical induction. The problem is that I am finding it hard to understand this type of proof as it is beyond my current level of understanding. I am looking for an answer that doesn't use mathematical induction. Could someone please explain in simple English (where possible) why 
$\bbox[yellow]{\displaystyle-\int\dfrac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\dfrac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x}
$
$\bbox[yellow]{\displaystyle=(-1)^m\int_{-1}^1\frac{\mathrm{d}^L}{\mathrm{d}x^L}(x^2-1)^{L}\frac{\mathrm{d}^L}{\mathrm{d}x^L}(x^2-1)^L\mathrm{d}x}$
I desperately need to understand this as this forms the final part of a $4$ step proof.
Thank you very much.
 A: We are trying to evaluate
$$\int_{-1}^1\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}dx$$
For $0\le n<m$. Let $$u=\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}$$
Then $$du=\frac{d^{\ell+1-m+n}}{dx^{\ell+1-m+n}}(x^2-1)^{\ell}dx$$
And $$dv=\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx$$
So $$v=\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}$$
Integrating by parts,
$$\begin{align}\int_{-1}^1u\,dv&=\left.u\,v\right|_{-1}^1-\int_{-1}^1v\,du\\
&=\int_{-1}^1\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx\\
&=\left.\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\right|_{-1}^1\\
&-\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\\
&=-\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\end{align}$$
Let's see... we were tasked with establishing that for any polynomial $p(x)$ and any $0\le n<k$ there is a polynomial $q(x)$ such that
$$\frac{d^n}{dx^n}\left((x-x_0)^kp(x)\right)=(x-x_0)^{k-n}q(x)$$
Well, it's true for $n=0$, and if true for some $0\le n<k-1$,
$$\begin{align}\frac{d^{n+1}}{dx^{n+1}}\left((x-x_0)^kp(x)\right)&=\frac d{dx}\frac{d^n}{dx^n}\left((x-x_0)^kp(x)\right)\\
&=\frac d{dx}\left((x-x_0)^{k-n}q(x)\right)\\
&=(k-n)(x-x_0)^{k-n-1}q(x)+(x-x_0)^{k-n}q^{\prime}(x)\\
&=(x-x_0)^{k-n-1}\left((k-n)q(x)+(x-x_0)q^{\prime}(x)\right)\\
&=(x-x_0)^{k-n-1}r(x)\end{align}$$
Where $r(x)$ is a polynomial. Having established the proposition for $n=0$ and having demonstrated its validity for $n+1$ given that it is true for $n<k-1$, it follows by mathematical induction for $0\le n<k$. From this we may conclude that
$$\left.\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\right|_{-1}^1=0$$
For $n<m$ because $\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}$ still has at least one of its $\ell$ factors of $(x^2-1)$ left in it.  
Using this result, we now are requested to show that for $0\le n\le m$,
$$\int_{-1}^1\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}dx\\=(-1)^n\int_{-1}^1\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx$$
Clearly it's true for $n=0$, and way back up at the top, we showed that if it's true for $0\le n<m$, then
$$\begin{align}&\int_{-1}^1\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}dx\\
&=(-1)^n\int_{-1}^1\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx\\
&=(-1)^n\left\{-\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\right\}\\
&=(-1)^{n+1}\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\end{align}$$
So it's true for $n+1\le m$, thus the result follows by mathematical induction and in particular for $n=m$,
$$\int_{-1}^1\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}dx=(-1)^m\int_{-1}^1\frac{d^{\ell}}{dx^{\ell}}(x^2-1)^{\ell}\frac{d^{\ell}}{dx^{\ell}}(x^2-1)^{\ell}dx$$
Thus we can see that
$$\int_{-1}^1\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}dx=(-1)^m\int_{-1}^1\left(2^{\ell}\ell!P_{\ell}(x)\right)^2dx=\frac{2\cdot2^{2\ell}(\ell!)^2}{2\ell+1}$$
Where we have used the Rodrigues formula for the Legendre polynomials
$$P_{\ell}(x)=\frac1{2^{\ell}\ell!}\frac{d^{\ell}}{dx^{\ell}}(x^2-1)^{\ell}$$
