Why it is said that second transistor cannot enter saturation mode in Darlington pair?
For an individual NPN bipolar junction transistor (BJT), the condition for saturation is

$$
V_E &lt; V_B &gt; V_C
$$

i.e., the transistor's base-emitter diode AND its base-collector diode must both be forward biased.

Applying Kirchoff's voltage law to the Darlington pair (fig. 1) we see that the voltage at its collector is

$$
V_{Q2.C} = V_{Q2.E} + V_{Q2.BE} + V_{Q1.CE}
$$

i.e., the voltage at Q2's collector \$V_{Q2.C}\$ is equal to the sum of Q2's emitter voltage \$V_{Q2.E}\$, plus the forward bias voltage across Q2's base-emitter diode \$V_{Q2.BE}\$, plus the voltage across Q1's collector-emitter path \$V_{Q1.CE}\$.



Figure 1.

Noting that

$$
V_{BE} := V_B - V_E
$$

by inspection, for \$V_{Q2.BE}\gt0\,V\$ and \$V_{Q1.CE}\gt0\,V\$ we see that Q2 transistor in a Darlington pair cannot meet the requirement for saturation:

$$
V_{Q2.E} &lt; V_{Q2.B} \stackrel{?}{&gt;} V_{Q2.C}
\\
\Rightarrow
V_{Q2.E} &lt; V_{Q2.B} \stackrel{?}{&gt;} (V_{Q2.E}+V_{Q2.BE}+V_{Q1.CE})
\\
\Rightarrow
V_{Q2.E} &lt; V_{Q2.B} \stackrel{?}{&gt;} (V_{Q2.E}+(V_{Q2.B}-V_{Q2.E})+V_{Q1.CE})
\\
\Rightarrow
V_{Q2.E} &lt; V_{Q2.B} \ngtr (V_{Q2.B}+V_{Q1.CE})
$$

Q2 approaches the transition point between forward active mode (small signal amplification) and soft saturation, but it never achieves saturation.