10. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=a_{n}+a_{n}^{2}\left(n \in \mathbf{N}^{*}\right)$. Let
$$
S_{n}=\frac{1}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)}, \quad T_{n}=\sum_{k=1}^{n} \frac{1}{1+a_{k}}
$$

Find the value of $S_{n}+T_{n}$.
Since $a_{n+1}=a_{n}+a_{n}^{2}$, we have $a_{n+1}=a_{n}\left(1+a_{n}\right)$, so $\frac{1}{1+a_{n}}=\frac{a_{n}}{a_{n+1}}$, thus
$$
S_{n}=\frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdot \frac{a_{3}}{a_{4}} \cdots \frac{a_{n}}{a_{n+1}}=\frac{a_{1}}{a_{n+1}}=\frac{1}{a_{n+1}}
$$

Since $a_{n+1}=a_{n}\left(1+a_{n}\right)$, we have $\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(1+a_{n}\right)}=\frac{1}{a_{n}}-\frac{1}{1+a_{n}}$, so $\frac{1}{1+a_{n}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}$, thus
$$
T_{n}=\sum_{k=1}^{n} \frac{1}{1+a_{k}}=\frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{2}}-\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n}}-\frac{1}{a_{n+1}}=\frac{1}{a_{1}}-\frac{1}{a_{n+1}}=1-\frac{1}{a_{n+1}}
$$

Therefore,
$$
S_{n}+T_{n}=1
$$