*EDIT (added some clarifications)*. The argument below provides a self-contained proof.

Introduce the shorthand $C^{-2}=A^{k+1}$. We need to show that
\begin{equation*}
\det(I+ CBABC) \ge \det(I + CABA^{-1}BAC).
\end{equation*}

We will prove this inequality by establishing the log-majorization
\begin{equation*}
\prod\nolimits_{j=1}^k\lambda_j(CABA^{-1}BAC)
\le
\prod\nolimits_{j=1}^k\lambda_j(CBABC),\quad 1\le j \le n.
\end{equation*}
This log-majorization is equivalent to $\lambda_1(\wedge^k (CABA^{-1}BAC)) \le \lambda_1(\wedge^k(CBABC))$ (for $1\le k\le n$). It suffices to show this for $k=1$; the general case follows similarly upon exploiting the multiplicativity of the wedge product.

Thus, to prove $\lambda_1(CABA^{-1}BAC) \le \lambda_1(CBABC)$, we use its *scale independence* and observe that for positive matrices $X$ and $Y$, we have
\begin{equation*}
[Y \le I \implies X \le I] \implies \lambda_1(X)\le \lambda_1(Y).
\end{equation*}

Thus, to prove the $\lambda_1$ inequality, it suffices to prove that
\begin{equation*}
CBABC \le I \Leftrightarrow\
\begin{bmatrix}
A^{k+1} & B\\
B & A^{-1}
\end{bmatrix} \ge 0\quad\implies CABA^{-1}BAC \le I.
\end{equation*}
But $\begin{bmatrix}
A^{k+1} & B\\
B & A^{-1}
\end{bmatrix} \ge 0$ only if $B\le \sqrt{A^{k+1}A^{-1}}=A^{k/2}$; and if this is so, then it must also be the case that $\begin{bmatrix}
A^{k-1} & B\\
B & A
\end{bmatrix} \ge 0$. Notice now that $A^{k-1} = C^{-2}A^{-2}$, and apply Schur complements to the latter matrix inequality to obtain
\begin{equation*}
BA^{-1}B \le C^{-2}A^{-2}\Leftrightarrow CABA^{-1}BAC \le I.
\end{equation*}
Thus, we have shown that $\lambda_1(CABA^{-1}BAC) \le \lambda_1(CBABC)$. In a similar manner we can prove the general case for $k>1$, which ends up establishing the desired log-majorization.

This finishes the proof of the claim because $\lambda(X) \prec_{\log} \lambda(Y) \implies \det(I+X) \le \det(I+Y)$.