How to deduce the formula for the $k$-th derivative of logarithm function? Given $\ln'(t) = \frac{1}{t}$. How Can we deduce the formula for the $k$-th derivative: $\ln^{(k)}(t) = \frac{(-1)^{k-1}(k-1)!}{t^k}$ for $k \geq 1$. I know how to prove this using induction, but how can just deduce this formula from the first derivative? Thanks.

By letting the $n$th derivative be denoted by the function $$f_n(x)=(\ln{(x)})^{(n)}\text{ with } f_0(x)=\ln{(x)}$$ We have that $$f_1(x)=x^{-1}$$ and every function following $f_1(x)$ is produced by multiplying the previous function by the current power and reducing the power by one. So we must have that $$f_n(x)=a_nx^{-n}$$ Where the constant terms $a_n$ are given by the recurrence relation $$a_1=1$$ $$a_n=-(n-1)a_{n-1}$$ This can easily be solved by noticing $$\begin{align} a_n &=-(n-1)a_{n-1}\\\ &=(-(n-1))(-(n-2))a_{n-2}\\\ &=(-1)^{n-1}(n-1)(n-2)\dots a_1\\\ &=(-1)^{n-1}(n-1)!\\\ \end{align}$$ So the functions are given by $$f_n(x)=(-1)^{n-1}(n-1)!x^{-n}$$