Relation between Pascal's Triangle and Euler's Number

First of all are you familiar with the Binomial Theorem? It is the reason why e.g. the row $14641$ looks like $(1.1)^4 = 1.4641$ -- just plug in $x = 1, y= 0.1$ into $(x+y)^n$.

Having said that, as you correctly pointed out, for later rows you need to "add zeros" to basically space out the numerals. This is equivalent to plugging in $x= 1, y = 0.01$ into $(x+y)^n$ for row $5$.

Now you also know that $\lim_{n \to \infty} (1 + 1/n)^n = e$; however this crucially relies on the $y = 1/n$ part of the sum dropping as $1/n$. If it drops faster or slower, the limit will be smaller than or larger than $e$. So in your "adding zeros" scheme it all depends on how fast you have to add zeros.

If your rule is always add just enough zeros so that the successive terms in a row do not "overlap", then even in the limit the first digit (the only digit before the decimal point) is always $1$, which is the left edge of the triangle. This rules out the limit being $e = 2.718...$ which has a leading digit $2$. In other words, following your rule, you would have to add zeros so fast (as a function of row number $n$) that the $y$ term in the sum drops faster than $1/n$.

In fact in your case $y = 10^{-D}$ where $D =$ no. of digits in the decimal representation of the biggest coefficient in the row, i.e. the center coefficient ${n \choose \lfloor n/2 \rfloor}$. This coefficient is known to grow exponentially fast, i.e. your $y$ would drop exponentially fast. As a further fact, your limit is actually $1$ because for really large $n$, you'd have to add soooo many zeroes that the first term (which is $n$) would make the decimal number look like $1.00000n...$

One of the definitions of $e$ is the limit

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$

Hence with $n=10^m$ you get better and better approximations $$(1+10^{-m})^{10^m}.$$

For instance

$$(1+0.0001)^{10000}=2.718146\cdots$$ where the first three decimals are exact.

On the other hand, this number is

$$\binom{10000}{0}+\binom{10000}{1}0.0001+\binom{10000}{2}0.00000001+\cdots\binom{10000}{10000}10^{-40000}$$ $$=1+1.0000+0.49995000+0.166616670000+\cdots$$

You indeed have the sum of Pascal's triangle entries with shifts, but the shifts are insufficient to separate the values and there are overlaps.

Compare to

$$(1+0.00000000001)^{10000}=1.00000010000000499950016661667\cdots$$