Convergence concerning the $\alpha$th derivative of $f(x)=x^\alpha-\alpha^x$ 
Let $x_\alpha$ be the stationary point of the $\alpha$th derivative of the function $f(x)=x^\alpha-\alpha^x$, and let $$\lambda_\alpha=\frac{f(-x_\alpha)}{-x_\alpha}.$$
For $\alpha\in\Bbb N$, does the limit $$\lim_{\alpha\to\infty}\bigg|\frac{\lambda_{\alpha+1}}{\alpha\lambda_\alpha}\bigg|$$ exist, and if so, what is its value?

We write $$\begin{align}f(x)=x^{\alpha}-\alpha^x&\implies f'(x)=\alpha x^{\alpha-1}-\alpha^x\ln\alpha\\&\implies\quad\quad\qquad\cdots\\&\implies f^{(\alpha)}(x)=\left[\prod_{i=0}^{\alpha-1}(\alpha-i)\right]-\alpha^x\ln^\alpha\alpha=0\end{align}$$ for stationary points. Then $$-x_\alpha=\frac1{\ln\alpha}\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]\tag{1}$$ so $$\begin{align}&\lambda_{\alpha}=\frac{\frac1{\ln^\alpha\alpha}\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]^\alpha-\frac{\ln^\alpha\alpha}{\prod_{i=0}^{\alpha-1}(\alpha-i)}}{\frac1{\ln\alpha}\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]}\\\implies&\lambda_{\alpha}=\frac{\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]^\alpha\prod_{i=0}^{\alpha-1}(\alpha-i)-\ln^{2\alpha}\alpha}{\ln^{\alpha-1}\alpha\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]\prod_{i=0}^{\alpha-1}(\alpha-i)}\tag{2}\end{align}$$ The expression for $|\lambda_{\alpha+1}/\alpha\lambda_{\alpha}|$ is extremely complicated so I will not give it here.
How should I continue? It seems that the limit does exist, but I would like a rigorous proof of it.
 A: Consider the following
$$f\left( x,z \right)={{x}^{z}}-{{z}^{x}}$$
Then it is not difficult to show
$$\frac{{{\partial }^{n}}}{\partial {{x}^{n}}}f\left( x,z \right)=\frac{\Gamma \left( z+1 \right)}{\Gamma \left( z-n+1 \right)}{{x}^{z-n}}-{{z}^{x}}\log {{\left( z \right)}^{n}}$$
Hence define 
$$\frac{\Gamma \left( z+1 \right)}{\Gamma \left( z-n+1 \right)}x_{z}^{z-n}-{{z}^{{{x}_{z}}}}\log {{\left( z \right)}^{n}}=0$$
Now consider the limit for $n\to z$
$$\Gamma \left( z+1 \right)-{{z}^{{{x}_{z}}}}\log {{\left( z \right)}^{z}}=0$$
This is the equation defining the stationary point ${{x}_{z}}$.  Solving for this and substituting into your definition for ${{\lambda }_{z}}$ we have after considerable algebra
$$\frac{{{\lambda }_{z+1}}}{z{{\lambda }_{z}}}=\frac{\log \left( z+1 \right)\log \left( z{{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)\left\{ {{\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{\log \left( z+1 \right)} \right)}^{z+1}}-\frac{{{\log }^{z+1}}\left( z+1 \right)}{\left( z+1 \right)\Gamma \left( z+1 \right)} \right\}}{z\log \left( z \right)\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)\left\{ {{\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}{\log \left( z \right)} \right)}^{z}}-\frac{{{\log }^{z}}\left( z \right)}{z\Gamma \left( z \right)} \right\}}$$ 
Now observe that $\lim\limits_{z\to\infty} \dfrac{{{\log }^{z}}\left( z \right)}{z\Gamma \left( z \right)}=0$, and many terms cancel in numerator and denominator (like for example the leading terms outside the curly brackets as they are of the same order in the limit).  We find therefore
$$\small\begin{align}\lim\limits_{z\to\infty} \frac{{{\lambda }_{z+1}}}{z{{\lambda }_{z}}}&=\lim\limits_{z\to\infty} {{\left( \frac{\log \left( z \right)\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{\log \left( z+1 \right)\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)} \right)}^{z}}\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)} \right)\\&=\lim\limits_{z\to\infty} {{\left( \frac{\log \left( z \right)+\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z \right) \right)}{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)} \right)}^{z}}\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)} \right)\\&=\lim\limits_{z\to\infty} {{\left( \frac{\log \left( z \right)}{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}+1 \right)}^{z}}\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)} \right)\end{align}$$
What remains are two limits: note
$$\lim\limits_{z\to\infty} \frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)}=\lim\limits_{z\to\infty} \left(\frac{1}{z}-\frac{\log \left( \log \left( z \right) \right)}{\log \left( z \right)}+\frac{\log \left( \Gamma \left( z \right) \right)}{z\log \left( z \right)}\right)$$ 
Now I think it is reasonable to assume that it is well known that $$\log \left( \Gamma \left( z+1 \right) \right)=\left( z-\tfrac{1}{2} \right)\log \left( z \right)-z+\mathcal O\left( 1 \right)$$ for large $z$ (use Stirling).  So we have
$$\lim\limits_{z\to\infty} \frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)}=\lim\limits_{z\to\infty} \frac{\left( z-\tfrac{1}{2} \right)\log \left( z \right)-z}{z\log \left( z \right)}=1$$
where we have used the fact that $\log \left( \log \left( z \right) \right)$ grows more slowly than $\log \left( z \right)$.  That’s one out of the way.  Next:
$$\lim\limits_{z\to\infty} \frac{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}{\log \left( z \right)}=1-\lim\limits_{z\to\infty} \frac{z\log \left( \log \left( z \right) \right)}{\log \left( z \right)}+\lim\limits_{z\to\infty} \frac{\log \left( \Gamma \left( z \right) \right)}{\log \left( z \right)}$$
Hence we have the asymptotic 
$$\lim\limits_{z\to\infty} \frac{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}{\log \left( z \right)}=1+\lim\limits_{z\to\infty} \frac{\left( z-\tfrac{1}{2} \right)\log \left( z \right)-z-z\log \left( \log \left( z \right) \right)}{\log \left( z \right)}\sim z$$ 
Hence
$$\lim\limits_{z\to\infty} {{\left( \frac{\log \left( z \right)}{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}+1 \right)}^{z}}\simeq \lim\limits_{z\to\infty} {{\left( \frac{1}{z}+1 \right)}^{z}}=e$$ 
Putting it all together we have shown therefore
$$\lim\limits_{z\to\infty} \bigg|\frac{{{\lambda }_{z+1}}}{z{{\lambda }_{z}}}\bigg|=e$$
