To be calculated $\lim _{n \rightarrow \infty}\left(\sqrt[3^{2}]{3} \cdot \sqrt[3^{3}]{3^{2}} \cdot \sqrt[3^{4}]{3^{3}} \ldots \sqrt[3^{n}]{3^{n-1}}\right)$.
We write the root quantities as powers of 3; in the product of powers with equal bases, we raise the common base (3) to the sum of the exponents. Thus, the sum

$$
S_{n-1}=\frac{1}{3^{2}}+\frac{2}{3^{3}}+\frac{3}{3^{4}}+\frac{4}{3^{5}}+\ldots+\frac{n-1}{3^{n}}
$$

arises. Now, we have

$$
\frac{1}{3} S_{n-1}=\frac{1}{3^{3}}+\frac{2}{3^{4}}+\frac{4}{3^{6}}+\ldots+\frac{n-2}{3^{n}}+\frac{n-1}{3^{n+1}}
$$

By subtracting the corresponding terms of the two equations:

$$
\begin{aligned}
& \frac{2}{3} S_{n-1}=\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\ldots+\frac{1}{3^{n}}+\frac{1}{3^{n+1}}-\frac{n}{3^{n+1}} \\
& 2 S_{n-1}=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots+\frac{1}{3^{n-1}}+\frac{1}{3^{n}}-\frac{n}{3^{n}} .
\end{aligned}
$$

The first $n$ terms on the right side form a geometric series with a common ratio of $\frac{1}{3}$. If we now let $n \rightarrow \infty$, then

$$
2 \lim _{n \rightarrow \infty} S_{n-1}=\frac{1}{3} \cdot \frac{1}{1-\frac{1}{3}}-\lim _{n \rightarrow \infty} \frac{n}{3^{n}}=\frac{1}{2}-\lim _{n \rightarrow \infty} \frac{n}{3^{n}}=\frac{1}{2}
$$

because

$$
\lim _{n \rightarrow \infty} \frac{n}{3^{n}}=0
$$

Indeed, the terms of the sequence

$$
\frac{1}{3}, \frac{2}{3^{2}}, \frac{3}{3^{3}}, \ldots \frac{10}{3^{10}}, \ldots \frac{n}{3^{n}}
$$

are continuously decreasing, so that if $n$ is sufficiently large, then $\frac{n}{3^{n}}$ is smaller than any given small number.

Therefore,

$$
\lim _{n \rightarrow \infty} S_{n-1}=\frac{1}{4}
$$

The limit of the product of the root quantities: $3^{\frac{1}{4}} = \sqrt[4]{3}$.