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How would one prove that the function $f(p)=p[(1+i)^p-1]$ is strictly decreasing? Here $i>-1$.

I've tried the usual approach, i.e. finding the derivative of $f$ and got a pretty terrible expression $$f'(p)=(1+i)^{1/p}(1-\frac 1p \log(1+i))-1$$ and I cannot find where the derivative is positive and where it's negative?

Any help would be appreciated.

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If $h$ is increasing and $g$ is increasing, then $h\cdot g$ is increasing if for all $x$, $h(x),g(x)> 0$. Indeed, if $x> y$, we have $h(x)\cdot g(x)>h(y)\cdot g(y)$ by the increasing properties.
Similarly, if $h$ is increasing and positive and $g$ is decreasing and negative, then $h\cdot g$ is decreasing by looking at it as $-(h\cdot -g)$ since $-g$ would be positive and increasing.

Assume now for the moment that we are working on the domains $(0,\infty)$ and take $h(p) = p$ and $g(p) = (1+i)^p - 1$.

If $i>0$, $g$ is increasing and positive. Thus $h\cdot g$ will be increasing. If $-1>i>0$, $g$ is decreasing and negative and thus $h\cdot g$ will be decreasing. If $i=0$, $g$ is zero and thus $h\cdot g$ will be zero.

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