$$(1-x^2)u''(x)-2xu'(x) + \ell(\ell+1)u(x)=0\tag{1}$$
Assume that $m$ is non-negative, differentiate $(1)$ (Legendre’s equation) $m$ times using Leibniz' theorem for differentiation to show that $$(1-x^2)v''(x)-2x(m+1)v'(x)+(\ell -m)(\ell + m + 1)v(x)=0\tag{*}$$ where $$v(x)\equiv \frac{d^mu}{dx^m}$$
From Leibniz' theorem the $m^{\text{th}}$ devivative of a product $fg$ is $$(fg)^m=f^{(m)}g+mf^{(m-1)}g^{(1)}+\cdots+\frac{m!}{n!(m-n)!}f^{(m-n)}g^{(n)}$$ where $f^{(m)}$ is the $m^{\text{th}}$ derivative of $f$, etc.
Hence, differentiating $(1)$ term by term,
$$\frac{d^m}{dx^m}\left[(1-x^2)u''\right]=(1-x^2)u^{(m+2)}-2mxu^{(m+1)}-\color{red}{2}m(m-1)u^{(m)}$$
I find the next term to be
$$\frac{d^m}{dx^m}\left[-2xu'\right]=-2xu^{(m+1)}-2mu^{(m)}$$
and the last term is
$$\frac{d^m}{dx^m}\left[\ell(\ell+1)u\right]=\ell(\ell+1)u^{(m)}$$
Substituting these into $(1)$ gives
$$(1-x^2)u^{(m+2)}-2xmu^{(m+1)}-\color{red}{2}m(m-1)u^{(m)}-2xu^{(m+1)}-2mu^{(m)} + \ell(\ell+1)u^{(m)}=0$$
which on simplification is
$$(1-x^2)v''(x)-2x(m+1)v'(x)+\left[\ell(\ell+1)-2m^2\right]v(x)=0$$
which is not the same as $(*)$.
In the solution the author gives, everything in my working is exactly the same except from the parts marked in red, according to the author's solution the $2$ should not be there (it should be unity). But the derivative of $2x$ is $2$. So I fail to understand why it should not be $2$.
Can someone please explain where I have gone wrong here?
$$\frac{m!}{n!(m-n)!}=\frac{m(m-1)}{2}$$ and this cancels the $2$ in red.