Prove that $g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))}$ is increasing in $x$, where $S_{n}(x)=\sum_{m=0}^{n}\frac{x^m}{m!}$ I want to prove that the function 
$$g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))},\,x >0$$ is increasing in $x$ for all $n$, where 
$ S_n(x)= \sum_{m=0}^{n}\frac{x^m}{m!}$. Differentiating gives something messy that I have not been able to prove it is non-negative. I have also been trying to find an appropriate $h(x)$ increasing and proving that $  h(g(x))$ is increasing $\ln(\cdot)$ seems a good candidate. 
By plotting I am almost convinced the statement is true.


Note: Here is a first step towards a complete answer. The point I like to address is the discrete log-concavity of the function $S_n$ for fixed $x>0$. This means $S_n$ fulfils
\begin{align*}
S_{n-1}^2(x)\geq S_n(x)S_{n-2}(x)\qquad\qquad x>0, n\geq 2\tag{1}
\end{align*}
A corresponding example can be found e.g. in the slides about Log-convexity and Log-concavity by Dmitry Karp. You might have a look at the section about Bessel functions where the inequality (1) for these functions is stated and some sharper results are presented afterwards.
Please note that in order to show $g(x)$ is increasing an even sharper inequality than (1) has to be proved. 

We start as @Arashium did. Using the differential operator $D_x$ we obtain
\begin{align*}
D_xS_n(x)=D_x\sum_{m=1}^{n}\frac{x^{m-1}}{(m-1)!}=\sum_{m=0}^{n-1}\frac{x^m}{m!}=S_{n-1}(x)
\end{align*}
and by omitting the argument $x$  we get
\begin{align*}
D_x\ln(S_n)&=\frac{1}{S_n}D_x(S_n)=\frac{S_{n-1}}{S_n}\\
\end{align*}
and
\begin{align*}
D_xg&=D_x\frac{\ln(S_n)}{\ln(S_{n-1})}
=\frac{\ln(S_{n-1})D_x\ln(S_n)-\ln(S_n)D_x\ln(S_{n-1})}{\ln^2(S_{n-1})}\\
&=\frac{1}{\ln^2(S_{n-1})}\left(\frac{S_{n-1}}{S_n}\ln(S_{n-1})-\frac{S_{n-2}}{S_{n-1}}\ln(S_n)\right)\tag{2}
\end{align*}

In order to show that the function $g$ is increasing we need $D_xg\geq 0$. 
Therefore the following is to prove according to (2)
  \begin{align*}
S_{n-1}^2\geq S_{n-2}S_n\frac{\ln(S_n)}{\ln(S_{n-1})}\tag{3}
\end{align*}

Here's the proof of the weaker inequality (1) showing that $S_n(x)$ is discrete log-concave for fixed $x>0$.
\begin{align*}
S_{n-1}(x)^2&-S_{n-2}(x)S_n(x)=\\
&=\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}\right)^2-S_{n-2}(x)\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}+\frac{x^{n}}{n!}\right)\\
&=S_{n-2}(x)\left(\frac{x^{n-1}}{(n-1)!}-\frac{x^{n}}{n!}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}S_{n-2}(x)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}\right)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=0}^{n-2}\frac{x^{m+1}}{m!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=1}^{n-1}\frac{x^{m}}{(m-1)!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}-\frac{x^{n-1}}{n(n-2)!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}\right)\\
&\qquad\qquad+\frac{x^{2n-2}}{(n-1)!(n-2)!}\left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{4}\\
&>0\\
\end{align*}

From the line (4) it's obvious that the inequality is valid and we may therefore conclude that $S_n$ is discrete log-concave.
The challenge is of course to sharpen the inequality (4) in order to obtain (3).

Note: Since the function $\ln$ is monotonically increasing we obtain due to the log-concavity of $S_n$
\begin{align*}
\ln\left(S_{n-1}^2(x)\right)&\geq\ln\left(S_n(x)S_{n-2}(x)\right)\\
2\ln\left(S_{n-1}(x)\right)&\geq\ln\left(S_n(x)\right)+\ln\left(S_{n-2}(x)\right)\\
\end{align*}
which may of some use for further calculations.
