If $a_n = \frac{{(n + 1)}^n}{n^{n - 1}} $ and $c > 0$ then $\lim_{n \to \infty} n^{1-c}(a^c_n-a^c_{n-1}) =ce^c $.
This is a generalization of How do you calculate this limit?, which is the case $c = \frac12$.
Here is my solution. As if often the case, the algebra is somewhat messy and I am interested in seeing if there is a more elegant solution.
$a_n =n\frac{{(n + 1)}^n}{n^{n }} =n(1+1/n)^n $.
We need the asymptotics of $(1+1/n)^{cn}$.
$\begin{array}\\ (1+1/n)^{cn} &=e^{cn\ln(1+1/n)}\\ &=e^{cn(1/n-1/(2n^2)+O(1/n^3))}\\ &=e^{c-c/(2n)+O(1/n^2))}\\ &=e^c\cdot e^{-c/(2n)+O(1/n^2))}\\ &=e^c\cdot (1-\frac{c}{2n}+O(1/n^2))\\ &=e^c-ce^c\frac{1}{2n}+O(1/n^2)\\ \end{array} $
Note: Wolfy says that $(1+1/n)^{cn} =e^c - ce^c\left(\frac{1}{2 n} - \frac{ 3 c + 8}{24 n^2} + \frac{c^2 + 8 c + 12}{48 n^3} \right) + O(1/n^4) $, which is comforting.
Therefore
$\begin{array}\\ a_n^c &=n^c(1+1/n)^{cn}\\ &=n^c(e^c-e^cc\frac1{2n}+O(1/n^2))\\ \end{array} $
and
$\begin{array}\\ a_{n-1}^c &=(n-1)^c(e^c-e^cc\frac1{2(n-1)}+O(1/n^2))\\ &=(n-1)^ce^c-e^cc\frac1{2(n-1)}+O(1/n^2))\\ \end{array} $
so that
$\begin{array}\\ a_n^c-a_{n-1}^c &=e^c(n^c-(n-1)^c)-\frac12 ce^c(\frac1{n}-\frac1{n-1})+O(1/n^2)\\ &=e^cn^c(1-(1-1/n)^c)-\frac12 ce^c(\frac1{n(n-1)})+O(1/n^2)\\ &=e^cn^c(1-(1-\frac{c}{n}+\frac{c(c-1)}{2n^2}))+O(1/n^2)\\ &=e^cn^c(\frac{c}{n}-\frac{c(c-1)}{2n^2})+O(1/n^2)\\ &=ce^cn^{c-1}-\frac{c(c-1)}{2n^{2-c}}+O(1/n^2)\\ \end{array} $
so that $n^{1-c}(a_n^c-a_{n-1}^c) =ce^c-\frac{c(c-1)}{2n}+O(1/n^{1+c}) \to ce^c $.
If $c = \frac12$, this is $\frac12 \sqrt{e}$ which agrees with the original problem.
I think the approach in my answer to the question which you have linked should work without much change.
To simplify typing I will use $A = a_{n}, B = a_{n - 1}$. Note that $A/n, B/n \to e$.
I first analyze $A^{c} - B^{c}$ and to that end we have \begin{align} A^{c} &= \exp(c\log A) = \exp\{c(n\log(n + 1) - (n - 1)\log n)\}\notag\\ &= \exp\{c(\log n + n\log(1 + 1/n))\}\notag\\ &= \exp(a)\text{ (say)}\notag \end{align} Similarly $$B^{c} = \exp\{c(\log n - (n - 2)\log(1 - 1/n))\} = \exp(b)$$ Note that $a - b \to 0$. And thus $$n^{1 - c}(A^{c} - B^{c}) = n^{1 - c}B^{c}\cdot\frac{\exp(a - b) - 1}{a - b}\cdot(a - b)$$ and thus it is sufficient to evaluate the limit of $$n^{1 - c}B^{c}(a - b) = (B/n)^{c}n(a - b)$$ Now $(B/n)^{c} \to e^{c}$ and $$n(a - b) = cn\{n\log(1 + 1/n) + (n - 2)\log(1 - 1/n)\}$$ which is same as $$c\{n^{2}\log(1 - 1/n^{2}) - 2n\log(1 - 1/n)\}$$ which tends to $c(-1 + 2) = c$. The final limit is thus $ce^{c}$. There is no need to use Taylor expansions. Just the limits $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ suffice.