Observing these terms finds that the difference of the two angles of each term equals $x$ which is a constant. Hence, it is attempting to employ the difference of tangents formula \myJustRefP{eq_tan_sum}. A further hint is that there is a term of $n$ in the right side of the target which can be sufficient to add $1$ to each of these $n$ terms on the left. \begin{align*} 1+\tan x\tan 2x &= \frac{\tan 2x - \tan x}{\tan(2x-x)}\\ 1 +\tan 2x\tan 3x&=\frac{\tan 3x - \tan 2x}{\tan(3x-2x)}\\ \cdots &=\cdots\\ 1 +\tan (n-1)x\tan nx&=\frac{\tan nx - \tan (n-1)x}{\tan(nx-(n-1)x)} \end{align*} Adding these $(n-1)$ equations together yields \begin{align*} &(n-1) + \tan x \tan 2x +\tan 2x \tan 3x +\cdots + \tan(n-1)x\tan nx \\ &=\frac{1}{\tan x}(\tan nx - \tan x) \end{align*} Rearranging the last equation leads to the desired claim.