123. Rationalize the denominator of the fraction:

$$
\frac{1}{\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{4}+\sqrt[5]{2}+1}
$$
$\triangle$ Let's use the identity

$$
a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)
$$

(see § 1, item 1.1). Let $a=\sqrt[5]{2}, b=1$; then $a^{5}-b^{5}=2-1=1$. Therefore, the conjugate factor for the denominator of the given fraction is $\sqrt[5]{2}-1$.

Answer: $\sqrt[5]{2}-1$.