A question about power amplification concept of a BJT
I think you are asking a very specific question about your own personal state of mind when faced with this text and I'll give a shot at addressing what I suspect is your misconception without hauling in a lot of abstract stuff:

The text is right when it says that the input current and the output current is the same. (Technically, it is not exactly the same. But it is the same "for all intents and purposes.") This is despite your recollection that the "output circuit current is beta times the base current." Your recollection is for a common-emitter arrangement. This is a common-base, though.

It might be easier for you to see this if I do a little diagram. Please don't worry about all the complexity present here. It's mostly just to get things arranged somewhat more realistically. It's simpler than it looks:







simulate this circuit &ndash; Schematic created using CircuitLab

Here, you have a common-base BJT arrangement. Think of the base as being "nailed to a voltage." Almost like you'd nail something to a wall (imagine the floor as ground and the ceiling as \$V_{CC}\$.) It's just suspended there.

The emitter "hangs off" from that nailed down base voltage, about one \$V_{BE}\$ less. The resistor \$R_E\$ is just there to set the DC operating current for \$Q_1\$, which will be about \$I_{E_Q}=\frac{V_B-V_{BE}}{R_E}\$. Also, \$I_{C_Q}=\frac{\beta}{\beta+1}I_{E_Q}\approx I_{E_Q}\$.

Now I want to call your attention to this last paragraph! Note that there is \$\beta\$ in the equation, just as you imagined when you wrote. Yes, \$\beta\$ is involved. But also note that because of the way things are arranged, it drops out, too. This is because the base can be ignored in this circuit, as the "important stuff" isn't going on at the base. It's going on at the emitter and the collector. And the emitter current and the collector current are almost equal to each other. This is why you read a statement telling you that the input current and the output current is the same (well, almost the same.)

Now let's look at the power situation. Suppose a tiny signal wiggles one side of transformer \$L_1\$. This will stimulate a wiggle on the other side, which acts like an AC signal source. The AC signal source here "sees" through the capacitor \$C_E\$ (which we assume looks like a dead-short at the frequency of interest) and causes a slight fluctuation at the emitter of \$Q_1\$. This tiny fluctuation modifies the standing (DC) emitter current slightly. The impedance seen by this signal source will be \$r_e=\frac{V_T}{I_{E_Q}}\$, due to the way that the BJT functions. (I'd have to take the derivative of its equation to show you why.) Here, \$V_T\approx 26\:\textrm{mV}\$ at room temperature and is called the thermal voltage. (It's a physical thing that is true for any matter and is based on the equi-partition rule of energy.) \$I_{E_Q}\$ is, of course, set by \$R_E\$. The output impedance will be \$R_L\$ (as seen through \$L_2\$.)

Let's do the equation thing now. I'll use small letters to signify the tiny AC signals and big letters to signify the DC values of things:

$$\begin{align*}
r_{in} &amp;= r_e &amp; \textrm{ } p_{in} &amp;= \frac{v_{in}^2}{r_e}\\\\
i_e &amp;= \frac{v_{in}}{r_e} &amp; \textrm{ } i_c &amp;= \frac{\beta}{\beta+1}i_e\approx i_e\\\\
p_{out} &amp;= i_c^2\: R_L = \left(\frac{\beta}{\beta+1}\frac{v_{in}}{r_e}\right)^2\: R_L\\\\
&amp;\therefore\\\\
\frac{p_{out}}{p_{in}} &amp;= \frac{\left(\frac{\beta}{\beta+1}\frac{v_{in}}{r_e}\right)^2\: R_L}{\frac{v_{in}^2}{r_e}} =\left(\frac{\beta}{\beta+1}\right)^2\: \frac{R_L}{r_e}\approx \frac{R_L}{r_e}
\end{align*}$$

As it turns out, this can be quite a large power gain! In fact, the common base arrangement can have probably the highest possible power gain for any single BJT arrangement.

Yet note that the emitter current (input side) is about the same as the collector current (output side.) And note that I did use \$\beta\$ here, too. It's just that it mostly "goes away" for purposes of figuring out what is happening with regard to the outputs and inputs of this arrangement.