A question on identification This question is about how to show identification of the fixed effects in a static panel linear model.
A1 (model): The model is
$$
Y_{it}=\alpha_i+X_{it}^\top \beta+\epsilon_{it}  
$$
for each $i=1,...,N$ and $t=1,...,T$, where $i$ indexes individuals and $t$ indexes time periods.
A2 (data): We assume to have an i.i.d. sample of $N$ observations $\{Y_{i1}, X_{i1},\dots, Y_{iT}, X_{iT}\}_{i=1}^N$ with $N$ large.
A3 (exogeneity): $E(\epsilon_{it}| X_{i1},..., X_{iT}, \alpha_i)=0$ for each $t=1,...,T$ and $i=1,...,N$.
Question:  In the so called "fixed effect model", $\alpha_1,..., \alpha_N$ are treated as parameters (together with $\beta$) and possibly estimated. How can we show that $(\alpha_1,..., \alpha_N, \beta)$ are identified under A1, A2, A3?
Remark: I think that $T$ large is also needed to identify $\alpha_1,..., \alpha_N$. Feel free to add this assumption.

My thoughts and doubts:
I have found several sources discussing how to estimate $(\alpha_1,..., \alpha_N, \beta)$ or how to identify/estimate $\beta$ alone (by differencing out $\alpha_1,..., \alpha_N$), but no papers or books explaining the joint identification of $(\alpha_1,..., \alpha_N, \beta)$.
I'm aware of the incidental parameter problem which prevents consistent estimation of $(\alpha_1,..., \alpha_N)$ when $T$ is fixed and  $N\rightarrow \infty$. Hence, I suppose that $(\alpha_1,..., \alpha_N)$  are not identified when $T$ is fixed and $N\rightarrow \infty$. The incidental parameter problem disappears if we also let $T\rightarrow \infty$. Does this imply that identification of  $(\alpha_1,..., \alpha_N)$  can be established? How?
In what follows,  I report my incomplete attempt.
$Y_i\equiv (Y_{i1},..., Y_{iT})$ and $X_i\equiv (X_{i1},..., X_{iT})$. $K$ is the size of $\beta$. I assume $NT>  N+K$.
First, I rewrite the model as
$$
Y_{it}=\sum_{\ell=1}^N\alpha_l 1\{i=\ell\}+X_{it}^\top \beta+\epsilon_{it}, 
$$
where I consider $\alpha_1,..., \alpha_N$ as parameters and the index $i$ as a random variable. Second, I rewrite A3
$$
E(\epsilon_{it}| i, X_{i1},..., X_{iT})=0, 
$$
for each $i=1,...,N$ and $t=1,...,T$.
By A3, for each   $i=1,\dots, N$, there exists a realisation  of the $T\times K $ matrix $X_{i}\equiv (X_{i1},..., X_{iT})$ (which I denote by $x_{i}\equiv (x_{i1},..., x_{iT})$) such that
$$
\begin{cases}
E(\epsilon_{i1}|i=1, X_{i} =x_1)=0 \\
\vdots\\
E(\epsilon_{iT}|i=1,  X_{i} =x_1 )=0 \\
\vdots\\
E(\epsilon_{i1}|i=N,  X_{i} =x_N )=0 \\
\vdots\\
E(\epsilon_{iT}|i=N, X_{i} =x_N )=0 \\
\end{cases} 
$$
In turn, by A1,
$$
\begin{cases}
E(Y_{i1}|i=1, X_{i} =x_1 )=\alpha_1+\beta x_{11} \\
\vdots\\
E(Y_{iT}|i=1, X_{i} =x_1 )=\alpha_1+\beta x_{1T} \\
\vdots\\
E(Y_{i1}|i=N, X_{i} =x_N )=\alpha_N+\beta x_{N1} \\
\vdots\\
E(Y_{iT}|i=N, X_{i} =x_N )=\alpha_N+\beta x_{NT} \\
\end{cases} 
$$
I can more compactly rewrite this system of equations as
$$
\underbrace{Y}_{NT\times 1}=\overbrace{\underbrace{\begin{pmatrix}
D & X
\end{pmatrix}}_{NT \times (N+K)}}^{\equiv \Gamma} \overbrace{\underbrace{\begin{pmatrix}
\alpha \\
\beta
\end{pmatrix}}_{(N+K)\times 1}}^{\equiv \phi}. 
$$
Next,
$$
\Gamma^\top Y= \Gamma^\top \Gamma \phi. 
$$
Thus,
$$
\phi=(\Gamma^\top \Gamma)^{-1}  \Gamma^\top Y 
$$
under the assumption that $\Gamma^\top \Gamma$ is invertible.
I would be done with the proof if I could claim that $Y$ is known for $N$ large under assumption A1. I don't think this is the case, though. I suppose that somehow we also need large $T$, but I don't see clearly how.
 A: If your $D$ is what I think it is, and taking into account the comments I made regarding the use of terminology and notation, then the additional assumption about $X$ I mentioned in the comments is just that $\Gamma^T\Gamma$ is invertible, which you figured out yourself. (It is not quite that trivial to find out what this means for your $X$ though. Large $N$ and $T$ may make it easier to fulfill this.)
What may be confusing is that you used $Y$ in the last equations for a vector that actually has the expected values of the $Y_{it}$ rather than the $Y_{it}$ themselves. The latter are random and not known, whereas the former are defined in the model framework. So you're basically done.
Note that for identifiability the $N$ and $T$ can be taken as fixed, however, as identifiability is about the model and not about a fixed sample, implicitly this assumes an infinite number of realisations of any $\epsilon_{it}$, or rather, more precisely and clearly, knowledge of the full distribution of $\epsilon_{it}$ (which carries all the randomness - I'm assuming that you treat the $X$ as fixed, otherwise you'd have to worry about identifying their distribution, too). This may look counterintuitive to you as you may interpret $N$ and $T$ as some kinds of sample sizes that potentially can go to $\infty$. In fact you need $T\to\infty$ (and some further conditions on $X$!) to have a consistent estimator (which as I wrote before is stronger than identifiability). Note that $N\to\infty$ will increase the number of $\alpha$-parameters, so not only will you have more observations at your disposal, you will also have to estimate more parameters, and this is not a standard identifiability problem. (It may however help with estimating $\beta$.)
