Example 10 Given $a>0$ and $a \neq 1$, the sequence $\left\{a_{n}\right\}$ is a geometric sequence with the first term $a$ and common ratio also $a$, let $b_{n}=$ $a_{n} \lg a_{n}(n \in \mathbf{N})$. If every term in the sequence $\left\{b_{n}\right\}$ is always less than the term that follows, find the range of values for $a$.

 untranslated portion: 

If every term in the sequence $\left\{b_{n}\right\}$ is always less than the term that follows, find the range of values for $a$.
Strategy: Isolate the parameter $k$ from $b_{k+1}-b_{k}>0$, and determine the range of $a$ through extremum calculation.

Solution: $\because a_{n}=a \cdot a^{n-1}=a^{n}$,
$$
\therefore b_{n}=a_{n} \lg a_{n}=n a^{n} \lg a \text {. }
$$

From $b_{k+1}-b_{k}=a^{k}[k(a-1)+a] \lg a>0$,
we get
$$
[k(a-1)+a] \lg a>0 .
$$

This implies $k>\frac{a}{1-a}$ for any natural number $k$.
Thus, $\frac{a}{1-a}<1$.

Therefore, $0<a<1$.