In the following we denote by $\left(x\frac{d}{dx}\right)^n f(x)$ the expression $x\frac{d}{dx}\left(x\frac{d}{dx}\left(\cdots x\frac{d}{dx}f(x)\right)\right)$ where the $x\frac{d}{dx}$ appears $n$ times. Also, $n,m,j$ always denote non-negative integers. By $\left|_{x=t}\right.$ we denote evaluation of an expression in the indeterminate $x$ (possibly involving derivatives) at $x=t$.
Claim 1. $\left(x\frac{d}{dx}\right)^n x^m(1+x)^j\left.\right|_{x=-1}=0$ for $j>n.$
Proof. We proceed by induction on $n$. For $n=0$ the claim is clear. Note $\begin{align*} \left(x\frac{d}{dx}\right)^n x^m(1+x)^j &=\left(x\frac{d}{dx}\right)^{n-1}x\frac{d}{dx} x^m (1+x)^j \\ &= m\left(x\frac{d}{dx}\right)^{n-1}x^m(1+x)^j+j\left(x\frac{d}{dx}\right)^{n-1}x^{m+1}(1+x)^{j-1} \end{align*} $ for $n\ge 1$. Evaluating at $x=-1$ and using the induction hypothesis gives the claim.
Claim 2. $(-1)^n=\sum^n_{j=1}(-1)^j{n+1\choose j+1}j^n$
Proof. By the binomial theorem we have $(t+1)^n=\sum^n_{j=0}{n\choose j}t^j$ for all $t\in\mathbb{R}$. Integrating on both sides with respect to $t$ from $0$ to $x\in\mathbb{R}\backslash\{0\}$ and dividing by $x$ gives
$x^{-1}F(x)=\sum^n_{j=1}{n\choose j}\frac{1}{j+1}x^j.$
where $F(x)=\int_0^x (t+1)^n dt$. Applying $\left(x\frac{d}{dx}\right)^n$ on both sides and noting that $\left(x\frac{d}{dx}\right)^n x^k=k^n x^k$ we get
$\left(x\frac{d}{dx}\right)^n x^{-1} F(x)=\sum^n_{j=1}{n\choose j}\frac{j^n}{j+1}x^j\;\;\;\;\;\;\;(*).$
Now we see by the fundamental theorem of calculus that
$\begin{align*} \left(x\frac{d}{dx}\right)^n x^{-1} F(x) &= \left(x\frac{d}{dx}\right)^{n-1} x\left(-x^{-2}F(x)+x^{-1}(1+x)^n\right)\\ &= -\left(x\frac{d}{dx}\right)^{n-1} x^{-1} F(x) + \left(x\frac{d}{dx}\right)^{n-1} (1+x)^n \end{align*} $ for $n\ge 1$. Evaluating at $x=-1$ gives $\left(x\frac{d}{dx}\right)^n x^{-1} F(x)\left|_{x=-1}\right.=-\left(x\frac{d}{dx}\right)^{n-1} x^{-1} F(x)\left|_{x=-1}\right.$ by Claim 1. Thus we have $\left(x\frac{d}{dx}\right)^n x^{-1} F(x)\left|_{x=-1}\right.=(-1)^{n+1} F(-1)=\frac{(-1)^n}{n+1}\;\;\;\;(**)$ where the last equality holds because $F(-1)=\int_0^{-1}(1+t)^ndt=-\int^1_0 t^n dt=\frac{-1}{n+1}.$ Evaluating $(*)$ at $x=-1$, plugging in $(**)$ and multiplying by $n+1$ gives $(-1)^n=\sum^n_{j=1}(-1)^j{n\choose j}\frac{n+1}{j+1}j^n=\sum^n_{j=1}(-1)^j{n+1\choose j+1}j^n $ as required.