The following argument is not a proof, but it may be converted to a proof.
We will write $0,1$ for the binomial coefficients taken modulo $2$, use tautologically these two colors, $0,1$ to color the coefficients.
The coefficients are considered as elements of the field $\Bbb F_2$.
Most of the time we use only its additive structure, i.e. the underlying abelian group structure $\Bbb Z/2$, but at some point i want to use the Frobenius morphism in characteristic two.
The line number zero in the Pascal triangle is simply
$$1$$
but it may be useful to think it as
$$
\dots\
0\
0\
0\
0\
0\
0\
1\
0\
0\
0\
0\
0\
0\
\dots
$$
or as
$$
\dots\
0\
0\
0\
0\
0\
0\
\color{red}{\blacktriangle}\
0\
0\
0\
0\
0\
0\
\dots
$$
It corresponds to $(1+x)^0$.
For $n=2$, $N=2^2$, we build the Pascal triangle over the field $\Bbb F_2$ up to row number $2^n-1=N-1=3$:
1
1 1
1 0 1
1 1 1 1
The last line is a line of ones, since the next line, line number $N=2^n$, is corresponding to the (coefficients in the) binomial expansion of $(1+x)^{2^n}=1^{2^n}+x^{2^n}$, so there survive only two extremal $1$ values, all other coefficients are zero. This works for a general $n$. Knowing line $N$, we have only one chance for the line $N-1$, it is a line of ones.
Now put a white $\nabla$ over the $0$ entry, some $\blacktriangle$ over the ones, so that the above may look like
$$\blacktriangle\\
\blacktriangle\nabla\blacktriangle
$$
and compare with the second triangle in
Sierpinski triangle evolution
In the computation of the next row, row number $2^2$, we have the line $1,0,0,0,1$, as already mentioned. Here is a picture:
1
1 1
1 0 1
1 1 1 1
1 0 0 0 1
And it is useful to think about it like:
$$
\color{red}{\blacktriangle}\\
1\ 1\\
1\ 0\ 1\\
1\ 1\ 1\ 1\\
\color{red}{\blacktriangle}\ 0\ 0\ 0\ \color{red}{\blacktriangle}
$$
Now we play again the "game of life", where a one bit gives life to the diagonally placed entries in the next row. Then the game works for the two $\color{red}{\blacktriangle}$ entries in the last row, as it worked for the upper $\color{red}{\blacktriangle}$, of course till they "interfere". We know exactly where is the interference, one step before the usage of the next Frobenius power, $(1+x)^8=1+x^8$, well, $(1+x)^7=\frac{1+x^8}{1+x}=\frac{1-x^8}{1-x}
=1+x+x^2+\dots+x^7$.
We have the following situation:
1
1 1
1 0 1
1 1 1 1
1 0 0 0 1
1 1 0 0 1 1
1 0 1 0 1 0 1
1 1 1 1 1 1 1 1
Now place a white $\nabla$ on the zero entries in the middle, so that the upper bar of $\nabla$ corresponds to the zero entries in the row $1\ 0\ 0\ 0\ 1$, and compare with the next picture in loc. cit. - we have reached the next stage. The next line is again a $1\ 0\ 0\ \dots\ 0\ 0\ 1$ line, and its picture is:
$$
\color{red}{\blacktriangle}\\
1\ 1\\
1\ 0\ 1\\
1\ 1\ 1\ 1\\
1\ 0\ 0\ 0\ 1\\
1\ 1\ 0\ 0\ 1\ 1\\
1\ 0\ 1\ 0\ 1\ 0\ 1\\
1\ 1\ 1\ 1\ 1\ 1\ 1\ 1\\
\color{red}{\blacktriangle}\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ \color{red}{\blacktriangle}
$$
The extremal $1$ values in the last row will suffer now the same copy+paste operation from the upper vertex, and we get a bigger $\nabla$ in the middle, we see the upper line of the white $\nabla$ in the last row, and the copy+paste game goes on...
The above inductive construction shows a self-similarity of the "picture" from one step ($N=2^n$) to the next one ($2N=2^{n+1}$), and it is natural to expect a self-similarity in the limit. (After a rigorous definition of the limit.) Note that the above construction goes parallelly to the construction of the Sierpinsky triangle.)
If it is really needed for a special purpose, i may try to become analytic, and establish an analytic formula for points, using their $2$-adic coordinates in the barycentric coordinates of the points inside the Sierpinski triangle - in the level $N$ and in the limit. (It is hard to write it down, and the proof of the convergence will not be intuitive.)
