why biasing an emitter follower with a transistor will cause the following transistor falling into active region?
You know that driving enough base current will fall the transistor in saturation region. 

But recall that for an NPN transistor, if \$V_C &lt; V_B &gt; V_E\$ --or simply, if \$V_B\$ is greater than both \$V_C\$ and \$V_E\$-- then the transistor will be in saturation region.

Let's test this in your circuit:

\$V_{C_{Q2}} = V_{CC} \$

Even if Q1 is saturated, \$V_{C_{Q1}}\$ will be between \$0\$ and \$V_{CC}\$. Because, as you can see, there's a voltage divider formed by R1, R2 and C-E resistance of Q1.

\$V_{B_{Q2}}=V_{C_{Q1}}\$, thus

\$0 &lt; V_{B_{Q2}} &lt;  V_{CC} \$ which means \$V_{B_{Q2}} &lt; V_{C_{Q2}}\$.

\$V_{B_{Q2}}\$ will never be greater than \$V_{C_{Q2}}\$ so Q2 cannot be in saturation region.