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 $ 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.
