38. Product. Simplify the following product:

$$
\left(3^{2^{0}}+1\right)\left(3^{2^{1}}+1\right)\left(3^{2^{2}}+1\right) \ldots\left(3^{2^{n}}+1\right)
$$
38. Multiplying the product by $1=\frac{1}{2}\left(3^{2^{0}}-1\right)$, we get

$$
\begin{gathered}
\frac{1}{2}\left(3^{2^{n+1}}-1\right), \quad \text { since } \quad\left(3^{2^{0}}-1\right)\left(3^{2^{0}}+1\right)=3^{2^{1}}-1 \\
\left(3^{2^{1}}-1\right)\left(3^{2^{1}}+1\right)=3^{2^{2}}-1 \quad \text { and so on. }
\end{gathered}
$$

In the general case, if we take any base $x>1$ instead of 3, our product will be $\frac{x^{2^{n+1}}-1}{x-1}$.

$[M . M ., 38,124($ March 1965).]