Use Cauchy Product to Find Series Representation of $[\ln(1 + x)]^2$ Problem:
Let $f(x) = [\ln(1 + x)]^2$.  Use the series for the logarithm to compute that
\begin{align*}
   f(x) =  [\ln(1 + x)]^2 = \sum_{n = 2}^{\infty}(-1)^n\Bigg(\sum_{k = 1}^{n - 1} \frac{1}{(n - k)k}\Bigg) x^n.
\end{align*}
Use this to evaluate the 5th derivative of $f$ evaluated at $0$.
My attempt:
Using $\ln(1 + x) = \sum_{k = 1}^{\infty} (-1)^{k - 1} \frac{x^k}{k}$, the Cauchy product of $f(x) = (\ln(1 + x))(\ln(1 + x))$ gives us
\begin{align*}
    f(x) &= \sum_{n = 1}^{\infty} \Bigg(\sum_{k = 1}^{n} \frac{(-1)^{n - k - 1}}{(n - k)} \Bigg(\frac{(-1)^{k - 1}}{k}\Bigg)\Bigg)x^n\\
         &= \sum_{n = 1}^{\infty} \Bigg(\sum_{k = 1}^{n} \frac{(-1)^{n - 2}}{(n - k)k} \Bigg)x^n\\
         &= \sum_{n = 1}^{\infty} \Bigg(\sum_{k = 1}^{n} \frac{(-1)^{n - 2}}{(n - k)k} \Bigg)x^n\\
         &= \sum_{n = 1}^{\infty} (-1)^{n}\Bigg(\sum_{k = 1}^{n} \frac{1}{(n - k)k} \Bigg)x^n.\\
\end{align*}
This has the form of the solution but I'm having trouble shifting the indices.  As for using it to find the 5th derivative, I just need to take the 5th derivative of the series and plug in $x = 0$?  Thanks for your help!

Note that for $-1<x< 1$ we have
$$\begin{align}
\log^2(1+x)&=\sum_{k=1}^\infty\sum_{m=1}^\infty\frac{(-1)^{k+m}x^{k+m}}{k\,m}\tag 1\\\\
&=\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^{n}x^{n}}{k\,(n-k)} \tag 2\\\\
&=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac{(-1)^{n}x^{n}}{k\,(n-k)} \tag 3\\\\
\end{align}$$
as was to be shown!

NOTES:
In going from $(1)$ to $(2)$, we introduced a new index $n=k+m$.  Then, with $m=n-k$, the lower limit for $n$ begins at $k+1$.
In going from $(2)$ to $(3)$, we interchanged the order of summation and note that $n\ge 2$ from $(2)$.
