The question boils down to the following statement.
Fact 1:
If $\lim_{k\rightarrow\infty}y_k =y\in\mathbb{R}$ and $\limsup_{k\rightarrow \infty}x_k=x\in\mathbb{R}$, then $\limsup_{k\rightarrow \infty}x_k^{y_k}=(\limsup_{k\rightarrow \infty} x_k)^{\lim_{k\rightarrow \infty}y_k}$
Fact 2:
Given $f:\mathbb{R}\mapsto \mathbb{R}$ and $\{x_k\}_{k\in\mathbb{R}}$ such that $\limsup_{k\rightarrow \infty}x_k=x\in\mathbb{R}$, if $f$ is continuous and increasing, then $\limsup_{k\rightarrow \infty} f(x_k) = f(\limsup_{k\rightarrow \infty} x_k)$.
Fact 2 is needed to prove Fact 1.
Proof of Fact 2,
Source:
Prove that $ \limsup f(x_n) = f(\limsup x_n) $
Show that $f(\sup(A))= \sup(f(A))$.
Since $f$ is continuous, and $\limsup_{k\rightarrow \infty}x_k=\lim_{k\rightarrow \infty} \sup_{n\geq k}x_n= x\in\mathbb{R}$, hence
\begin{equation}
\lim_{k\rightarrow \infty} f(\sup_{n\geq k}x_n)=f(\lim_{k\rightarrow \infty} \sup_{n\geq k} x_n)=f(x)
\end{equation}
Moreover, since $f$ is increasing, then for $k\in\mathbb{N}$ large enough, $f(\sup_{n\geq k}x_n)\geq f(x_n)$ for all $n\geq k$, thus \begin{equation}f(\sup_{n\geq k}x_n)\geq \sup_{n\geq k} f(x_n)\end{equation}
In other hand, let $z_k = \sup_{n\geq k} x_n\in\mathbb{R}$ for $k$ large enough, for every $m\in\mathbb{N}$, there exists $x_{n_m} \in \{x_n: n\geq k\}$ such that $z_k-\frac{1}{m}<x_{n_m}\leq z_k$. By Squeeze theorem, $\lim_{m\rightarrow \infty} x_{n_m} = z_k$. Since $f$ is continuous, then $\lim_{m\rightarrow \infty} f(x_{n_m})=f(z_k)$.
Therefore, $\forall\,\varepsilon>0$, there exists $M\in \mathbb{N}$ such that $\forall\,m\geq M$, $|f(x_{n_m}) - f(z_k)|<\varepsilon$. So, $\forall\,m\geq M$,
\begin{equation}
f(z_k) = f(z_k) - f(x_{n_m}) + f(x_{n_m}) \leq |f(z_k) - f(x_{n_m})| + f(x_{n_m})<\varepsilon+f(x_{n_m})\leq\varepsilon+\sup_{n\geq k} f(x_n)
\end{equation}
Hence, $\forall\, \varepsilon >0$, $f(\sup_{n\geq k}x_n)<\varepsilon + \sup_{n\geq k} f(x_n)$. Therefore, \begin{equation}f(\sup_{n\geq k}x_n) \leq \sup_{n\geq k} f(x_n)\end{equation}
Now, we have $f(\sup_{n\geq k}x_n) = \sup_{n\geq k} f(x_n)$.
Consequently, from the first equation, we have
\begin{equation}
\lim_{k\rightarrow \infty}f(\sup_{n\geq k}x_n) = \lim_{k\rightarrow \infty} \sup_{n \geq k} f(x_n) = \limsup_{k\rightarrow \infty} f(x_k) = f(\limsup_{k\rightarrow \infty} x_k)
\end{equation}
Q.E.D.
Proof of Fact 1,
Source:
$\limsup_{n\to\infty}{a_n}^{b_n}= \limsup_{n\to\infty}{a_n}^{\lim\limits_{n\to\infty}b_n}$ under certian conditions.
lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $
Notice $e^x$ and $ln(x)$ are increasing and continuous, use Fact 2 in the proof.
\begin{align}
(\limsup_{k\rightarrow \infty} x_k)^{\lim_{k\rightarrow \infty} y_k} & = x^{\lim_{k\rightarrow \infty} y_k}\\
& = e^{ln(x) \lim_{k\rightarrow \infty} y_k}\\
& = e^{(\lim_{k\rightarrow \infty} y_k) ln(\limsup_{k\rightarrow \infty} x_k)}\\
& = e^{(\lim_{k \rightarrow \infty} y_k) \limsup_{k \rightarrow \infty} ln(x_k)}\\
& = e^{\limsup_{k \rightarrow \infty} y_k ln(x_k)}\\
& = \limsup_{k \rightarrow \infty} e^{y_k ln(x_k)}\\
& = \limsup_{k \rightarrow \infty} x_k^{y_k}
\end{align}
Because of $f(x) = ln(x)$ and $\limsup_{k \rightarrow \infty} x_k = x \in \mathbb{R}$, the fourth equality is implied by Fact 2. The fifth equality is implied by lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $. Moreover, due to $g(x) = e^x$ and $\limsup_{k\rightarrow \infty} y_k ln(x_k) \in \mathbb{R}$, the sixth equality is also implied by Fact 2. This finishes the proof.
Q.E.D
In this post, given $\limsup_{k\rightarrow \infty} |a_k|^{\frac{1}{k}}\in \mathbb{R}$ and $\lim_{k \rightarrow \infty} \frac{k}{k-1}$, using $|a_k|^{\frac{1}{k-1}} = |a_k|^{\frac{1}{k} \frac{k}{k-1}}$, the result $\limsup_{k\rightarrow \infty}|a_k|^{\frac{1}{k}} = \limsup_{k\rightarrow \infty}|a_k|^{\frac{1}{k} \frac{k}{k-1}}$ is a direct application of Fact 1.
