Can someone help me to confirm this identity that I have established, I really have no idea how I would go about proving that this is true.
By the way, this is not a homework assignment, I am just genuinely curious. Thank you!
If $f(x)=x^x$, then
$$f^{(n+1)}(x)=f^{(n)}(x)(\ln(x)+1)+\sum_{k=1}^n (-1)^{k+1}\frac{n!}{k(n-k)!}f^{(n-k)}(x)$$ $$\text{ for } n\in\mathbb{N}$$
where $f^{(n)}(x)$ is the $n^{\text{th}}$ derivative of $f(x)$
For $n = 1$,
$$\begin{align*} f(x) &= x^x\\ & = e^{x\ln x}\\ f'(x) &= \frac{d}{dx} e^{x\ln x}\\ &= e^{x\ln x} \left(\ln x + 1\right)\\ &= x^x\left(\ln x + 1\right)\\ \end{align*}$$
Let $g(x) = \ln x + 1$. Then for $k \ge 1$, $$g^{(k)}(x) = (-1)^{k-1}(k-1)!\ x^{-k}$$
From the second derivative onwards, i.e. $n \ge 1$, using the general Leibniz rule,
$$\begin{align*} f^{(n+1)}(x) &= \frac{d^n}{dx^n}f'(x)\\ &= \frac{d^n}{dx^n}\left[f(x)\ g(x)\right]\\ &=\sum_{k=0}^{n}\binom{n}{k}\ f^{(n-k)}(x)\ g^{(n)}(x)\\ &= f^{(n)}(x)\ g^{(0)}(x) + \sum_{k=1}^{n}\binom{n}{k}\ f^{(n-k)}(x)\ g^{(n)}(x)\\ &= f^{(n)}(x) \left(\ln x + 1\right) + \sum_{k=1}^{n}\binom{n}{k}\ f^{(n-k)}(x)\left[(-1)^{k-1}(k-1)!\ x^{-k}\right]\\ &= f^{(n)}(x) \left(\ln x + 1\right) + \sum_{k=1}^{n}(-1)^{k+1}\frac{n!}{k(n-k)!}\ f^{(n-k)}(x)\ \color{red}{x^{-k}}\\ \end{align*}$$
The red $x^{-k}$ is what I think the question may be missing, but I could be wrong.
I recently found something related. Given that $f(x)$ and $g(x)$ are continuous and $n$-times differentiable on some interval $I\subseteq \Bbb R$, and $u(x)=f(x)g(x)$, then $$u^{(n)}(x)=\sum_{k=0}^{n}{n\choose k}f^{(n-k)}(x)g^{(k)}(x).$$
If you can find $n$-times differentiable $f(x)$ and $g(x)$ such that $x^x=f(x)g(x)$, then you have your theorem.
In Theorem 3 of the paper [1] below, it was obtained that \begin{multline}\label{x-poer-x-thm3-driv}\tag{QEq} \frac{\operatorname{d}^n}{\operatorname{d}x^n}\bigl[(1+x)^{t(1+x)}\bigr] =n!(1+x)^{t(1+x)-n}\\ \times\sum_{k=0}^{n}t^k(1+x)^{k} \sum_{j=0}^{k} \Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr] \frac{[\ln(1+x)]^{k-j}}{(k-j)!}. \end{multline} Replacing $1+x$ by $x$ in \eqref{x-poer-x-thm3-driv} yields \begin{equation}\label{x-poer-x-thm3-driv-repl} \frac{\operatorname{d}^n}{\operatorname{d}x^n}\bigl(x^{tx}\bigr) =n!x^{tx-n}\sum_{k=0}^{n}t^kx^{k} \sum_{j=0}^{k} \Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr] \frac{(\ln x)^{k-j}}{(k-j)!}. \end{equation} See also related comments on the answer at the web site https://mathoverflow.net/a/439188.
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