How to use series to prove this inequality? $\varphi(x, p) = \frac 1p (e^{px}-1)$ is increasing in $p$ for $p > 0$. I am trying to show that for any fixed $x$, the function
$$g(p) = \frac 1p (e^{px}-1)$$
is increasing on $(0, \infty)$. To do this, I found the derivative
$$g'(p) = e^{px} \left(\dfrac {x}{p} - \dfrac {1}{p^2}\right) + \dfrac {1}{p^2}$$
Solving $(\frac {x}{p} - \frac {1}{p^2}) > 0$ for $p$ yields $p > 1/x$. Thus, for $p > 1/x$, we have $g'(p) > 0$. However, I'm not sure what to do for $p \le \dfrac 1x$.
Since we have a mix of transcendental and algebraic functions, I believe now is the time to approximate $e^{px}$ by a Taylor polynomial. However, I am not sure how to do that.
Any help is appreciated! I welcome any solution, but I am especially interested in a solution using Taylor series.

All is consequence of the convexity of  $\phi(t)=e^{tx}$ ($x$ fixed). Since this type of arguments appear all over the place, I am taking the liberty of explaining this a little further:
Recall that  $\varphi:(a,b)\rightarrow\mathbb{R}$,
$-\infty\leq a<b\leq \infty$, is convex if
$$\begin{align}
\varphi((1-t) x+ t y)\leq (1-t)\varphi(x)+t \varphi(y)\tag{1}\label{convex}
\end{align}$$
for any $a<x<y<b$ and $0\leq t\leq 1$. If strict inequality holds
in $\eqref{convex}$ with $0<t<1$, then $\varphi$ is strictly convex.
Geometrically, if $\varphi$ is convex and $a<x<u<y<b$ then the point
$(u,\varphi(u))$ on the graph of $\varphi$ lies below the straight line
joining  $(x,\varphi(x))$ and $(y,\varphi(y))$. Let   $u=(1-t)x+ty$,
It is easy to check that
$\eqref{convex}$ is equivalent to any of the inequalities
$$
\begin{align}
\frac{\varphi(u)-\varphi(x)}{u-x}\leq\frac{\varphi(y)-\varphi(x)}{y-x}\leq 
\frac{\varphi(y)-\varphi(u)}{y-u}\tag{2}\label{convex-equiv}
\end{align}
$$
For fixed  $a<x<b$, inequalities~\eqref{convex-equiv} show that the map
$u\mapsto \tfrac{\varphi(u)-\varphi(x)}{u-x}$
decreases as $u\searrow x$  and  increases as $u\nearrow x$.
In your case
$$
\frac 1p (e^{px}-1)=\frac{\phi(p)-\phi(0)}{p-0}
$$
