You're asking to prove there's no positive integer $C$ and an infinite set of distinct positive integers $a_i$ such that
$$(a_{k+1})^k \mid C^k\left(\prod_{i=1}^{k}a_i\right) \tag{1}\label{eq1A}$$
is true for all positive integers $k$. Assume there is some such positive integer $C$ and set of $a_i$. For any prime $p_j$ which divides $C$ or one of the $a_i$, using the $p$-adic order function, let
$$b_{j,i} = \nu_{p_{j}}(a_i), \; \; c_j = \nu_{p_{j}}(C) \tag{2}\label{eq2A}$$
We'll prove using induction, with the harmonic numbers (i.e., $H_1 = 1$ and $H_{n+1} = H_n + \frac{1}{n+1} \; \forall \; n \ge 1$), for all $k \ge 1$ that
$$kb_{j,k+1} \le kc_j + \sum_{i=1}^{k}b_{j,i} \le kH_{k}c_j + kb_{j,1} \; \; \to \; \; b_{j,k+1} \le H_{k}c_j + b_{j,1} \tag{3}\label{eq3A}$$
For $k = 1$, we get from \eqref{eq1A} and \eqref{eq2A} that
$$b_{j,2} \le c_{j} + b_{j,1} = H_{1}c_{j} + b_{j,1} \tag{4}\label{eq4A}$$
so \eqref{eq3A} works in the base case since $H_1 = 1$. Assume \eqref{eq3A} is true for $k = n$ for some $n \ge 1$. Then, for $k = n + 1$, \eqref{eq1A}, \eqref{eq2A} and \eqref{eq3A} give that
$$\begin{equation}\begin{aligned}
(n+1)b_{j,n+2} & \le (n + 1)c_j + \sum_{i=1}^{n+1}b_{j,i} \\
& = \left(nc_j + \sum_{i=1}^{n}b_{j,i}\right) + c_j + b_{j,n+1} \\
& \le (nH_{n}c_j + nb_{j,1}) + c_j + (H_{n}c_j + b_{j,1}) \\
& = ((n+1)H_{n} + 1)c_j + (n + 1)b_{j,1} \\
& = (n + 1)\left(H_{n} + \frac{1}{n+1}\right)c_j + (n + 1)b_{j,1}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Dividing both sides of \eqref{eq5A} by $n+1$ gives
$$b_{n+2} \le \left(H_{n} + \frac{1}{n+1}\right)c_j + b_{j,1} = H_{n+1}c_j + b_{j,1} \tag{6}\label{eq6A}$$
so \eqref{eq3A} holds for $k = n + 1$ as well. Thus, by induction, \eqref{eq3A} holds for all positive integers $k$.
Note \eqref{eq3A} shows all prime factors of any $a_i$ must divide $Ca_1$, so there are only a finite number of them, say some $r \ge 1$. Also, the harmonic numbers being a strictly increasing sequence means that $b_{j,i} \le H_kc_j + b_{j,1}$ for all $1 \le i \le k + 1$. Thus, considering all of the $r$ distinct prime factors, we have that all $a_i$ for $1 \le i \le k + 1$ must be distinct positive factors of
$$M = \prod_{j=1}^{r}p_j^{e_j}, \; e_j = \lfloor H_{k}c_{j} \rfloor + b_{j,1} \tag{7}\label{eq7A}$$
Thus, using the number-of-divisors function $\sigma_0()$, the harmonic number upper bound of $H_k \le \ln(k) + 1$ and $c_{m} = \max(c_{1} + 1, \ldots, c_{r} + 1)$, we get for large enough $k$ (i.e., where $\ln k \gt \max(c_{1} + b_{1,1} + 1, \ldots, c_{r} + b_{r,1} + 1)$) that
$$\begin{equation}\begin{aligned}
k + 1 & \le \sigma_0(M) \\
& = \prod_{j=1}^{r}(\lfloor H_{k}c_{j} \rfloor + b_{j,1} + 1) \\
& \le \prod_{j=1}^{r}(c_{j}\ln{k} + c_{j} + b_{j,1} + 1) \\
& \lt \prod_{j=1}^{r}((c_{j} + 1)\ln{k}) \\
& \le \prod_{j=1}^{r}(c_{m}\ln{k}) \\
& = c_{m}^{r}(\ln{k})^{r}
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
With $k = e^x$, we get $k \to \infty$ means $x \to \infty$. Also, \eqref{eq8A} becomes
$$e^x + 1 \le c_{m}^{r}x^{r} \tag{9}\label{eq9A}$$
However, exponential functions grow faster than any polynomial (e.g., as can been seen here by using L'Hôpital's rule $r$ times to get $\lim_{x \to \infty}\frac{e^x + 1}{c_{m}^{r}x^{r}} = \lim_{x \to \infty}\frac{e^x}{c_{m}^{r}(r!)} = \infty$, or by using a post like Proving exponential is growing faster than polynomial), so \eqref{eq9A} can't hold for large enough $x$ and, thus, large enough $k$.
This contradiction concludes proving that \eqref{eq1A} can't always be true.
