Why is noncentral t-distribution for sample size determination for small sample? For sample size determination for small sample, population distribution of which is normal, the test statistic $$T = \frac{\bar{x}-\mu_0}{S/\sqrt{n}}, ~~~\text{where $\mu_0$ is null value}$$
is considered under the assumption of the alternative hypothesis $\mu = \mu' \gt \mu_0$.
Since $T$ follows noncentral $t$ distribution, chart or table is used to determine sample size $n$ with desired type II error.
The above is typical explanation available in statistics text books.
Now I'm curious what's wrong with the following reasoning:
If we assume that the alternative hypothesis $μ=μ′$ is correct $T = \frac{\bar{x} - \mu'}{S/\sqrt{n}}$ should follow $t$-distribution by the same argument that $T = \frac{\bar{x} - \mu_0}{S/\sqrt{n}}$ follows $t$-distribution when  $μ=μ_0$. Then we can proceed on to calculate
$$
Pr(\frac{\bar{x}-\mu'}{S/\sqrt{n}} \leq \frac{\mu_0 + t_{\alpha, n-1}s/\sqrt{n} - \mu'}{s/\sqrt{n}}) = \beta(\mu')~~~\text{(*)}
$$
to determine sample size n that satisfies the required $\beta$.
noncentral t-distribution cannot be involved at all.
Specifically, if my assumption that $(\bar{x}-\mu')/(s/\sqrt{n})$ follows t distribution is correct, we can simply set:
$$
\frac{\mu_0 + t_{\alpha, n-1}s/\sqrt{n} - \mu'}{s/\sqrt{n}} = t_{\beta(\mu')}
$$
and solve for $n$. No need to summon noncentral t distribution, no need to use $\beta$ curve to find $n$
=================================================================
With help of @BruceET, I found my silly mistake.
The equation to determine a rejection region is
$$
Pr(\frac{\bar{x} - \mu_0}{S/\sqrt{n}} > t_{\alpha, n-1}) = \alpha
$$
From this I wrongly thought the rejection region in terms of $\bar{x}$ is
$$
\bar{x} > \mu_0 + t_{\alpha, n-1}\frac{s}{\sqrt{n}}
$$
This is wrong because $s$ itself is a random variable and therefore cannot be included in the bounds. If it were, I could have argued the eq.(*) could be used to calculate power for the alternative hypothesis.
 A: Let $X_1, \dots, X_n$ be a random sample from $\mathsf{Norm}(\mu, \sigma).$ Suppose you want to test $H_0: \mu=10$ against $H_a: \mu > 10$ at the 5% level of significance when $n = 16.$
Then the t statistic for this test is
$T_0 = \frac{\bar X - 10}{S/\sqrt{n}}$ and $H_0$ is rejected if $T_0 \ge t^* = 1.753,$ where $t^*$ cuts
probability 5% from the upper tail of Student's t
distribution with $\nu =n-1 = 15$ degrees of freedom:
$\mathsf{T}(\nu=15).$ Computation in R.
qt(.95, 15)
[1] 1.75305

The power $\pi(12) = P(\mathrm{Rej}\, H_0\, |\, \mu = 12)$ against the specific alternative $\mu_a = 12$ is
$$\pi(\mu_a) = \pi(12) = P\left(\frac{\bar X - \mu_0}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)
= P\left(\frac{\bar X - \mu_a + (\mu_a - \mu_0)}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)\\
= P\left(\frac{\bar X - \mu_a + 2}{S/\sqrt{n}}\ge t^*\,|\mu_a=12\right)
= P\left( \frac{ Z + \delta }{ \sqrt{V/\nu} } \ge t^* \right),$$
where $\delta = \sqrt{n}(\mu_a-\mu_0)/\sigma = 2\sqrt{n}/\sigma,\,$
$Z \sim\mathsf{Norm}(0,1),$ and $V = \nu S^2/\sigma^2
\sim \mathsf{Chisq}(\nu).$
That is, by definition,  $\frac{ Z + \delta }{ \sqrt{V/\nu}}\sim \mathsf{T}(\nu, \delta),$ Student's noncentral t distribution with $\nu$ degrees of freedom and noncentralty parameter $\delta.$
In particular, for our example, $\delta = 2\sqrt{16}/\sigma.$
For $\sigma = 2,$ we have $\delta = 4$ and $\pi(12) = 0.985.$
1 - pt(qt(.95, 15), 15, 4) 
[1] 0.9848477

The following simulation in R gives essentially the same power. With a million iterations one
can expect three-place accuracy.
set.seed(1128)
pv = replicate(10^6,  t.test(rnorm(16, 12, 2), 
               mu=10, alt="g")$p.val)
mean(pv <= .05)
[1] 0.984809

Ref: Except for notation, parts of the above parallel Sect 2.3 of Bain & Englehardt (1992), Intro. to Probability and Math. Stat., Duxbury, p400.
