We have $$\frac {n^{100}}{1.01^{n}} = \left(\frac{n}{1.01^{n/100}}\right)^{100}
$$ By Bernoulli's inequality on the inside quantity, $$\begin{align}\left(\frac{n}{\left(1+\frac{1}{100}\right)^{n/100}}\right)^{100}
&\leq \left(\frac{n}{1 + \frac{1}{100}\cdot\frac{n}{100}}\right)^{100} \\ &< \left(\frac{n}{\frac{n}{10000}}\right)^{100} \\ &= 10000^{100}
\end{align}$$
The issue is that all sequences are implicitly infinite in this context. Sometimes in my texts, what you write $(a_n)$ in your post, I would see written as $(a_n)_{n\in\Bbb N}$ in my work. We only drop the "$n\in\Bbb N$" bit when it's understood what is going on (that it's infinite).
That is,
$$(a_n)_{n\in\Bbb N} = (a_1, a_2, a_3, a_4, ... \text{and so on and so forth, forever neverending})$$
Of course, then, by convention, we have that the sequence $n_1,n_2,n_3,...$ (where each term is strictly greater than the last) which defines the indices of the subsequence is also infinite. That is to say, every sequence (in this context) is infinite, as are all subsequences.
It is in fact that infiniteness that allows us to conclude that each subsequence of a convergent sequence converges to the same limit as the sequence itself - as you have seen, it doesn't hold in the finite case.
Also, for what it's worth, this theorem is discussed on MSE here.
Best Answer
An idea:
$$\left(1+\frac1{n^2+2n}\right)^{n^2}=\left(1+\frac1{n^2+2n}\right)^{n^2+2n}\left(1+\frac1{n^2+2n}\right)^{-2n}\xrightarrow[n\to\infty]{}e\cdot1=e$$