You just need to note that $(1+x)^5$ has a full binomial expansion given by $$
(1+x)^5= 1+5x+10x^2+10x^3+5x^4+x^5
$$
Therefore, if the approximation $(1+x)^5 \approx 1+5x$ does not work, then try $1 + 5x + 10x^2$ instead. This leads to $1.051$ as an approximation. If necessary, going one further down the expansion $1.05101$, which upon substitution will not change the answer much. This is the stipulation of some comments in this thread as well : the answer will stabilize near $2967$ by the second order approximation.
In the absence of division
Suppose, however, you wanted to completely avoid division : your calculator hates division (or you hate it). Then, you need to focus on the function $\frac{x^5}{x^5-1}$ and how it behaves near $1$, because $1.01$ is close to $1$.
We write it as $\frac{x^5}{x^5-1} = 1 + \frac{1}{x^5-1}$ so that we only need to focus on $\frac{1}{x^5-1}$. However, there's a problem : we cannot "Taylor expand" $\frac{1}{x^5-1}$ around the point $1$, because it goes to $+\infty$ as $x \to 1$. However, we can still isolate the "bad" part by finding the rate at which $\frac{1}{x^5-1}$ goes to $+\infty$ as $x$ goes to $1$. Then, the remaining part will admit a Taylor expansion.
To do that, observe that
$$
\frac{1}{x^5-1} = 1+\frac{1}{(x-1)(1+x+x^2+x^3+x^4)} \approx \frac{1}{5(x-1)}
$$
Basically, $\frac{1}{x^5-1}$ behaves "like" $\frac{1}{5(x-1)}$ as $x$ is closed to $1$. We are led to expect that removing the "bad" part $\frac{1}{5(x-1)}$ from $\frac{1}{x^5-1}$ should lead to something that is Taylor expandable around $1$.
You will observe that the approximation $\frac{1}{x^5-1} \approx \frac{1}{5(x-1)}$ is not good enough for the question you're solving (because you'll get $144 \times 21 = 3024$ which is not good enough).
As I said, we need to see if removing the "bad" part $\frac{1}{5(x-1)}$ from $\frac{1}{x^5-1}$ leads to something that is finite around $x \to 1$, so that it can be Taylor expanded if necessary. That's why we consider
$$
\lim_{x \to 1} \frac{1}{x^5-1} - \frac{1}{5(x-1)} = \frac{5x-x^5-4}{5(x-1)(x^5-1)}
$$
A couple of L'Hospitalizations (quite easy ones, because the product below is easy to expand) later, you land at the quantity $-\frac{2}{5}$. This will tell you that the function $$
g(x) =\frac{1}{x^5-1} - \frac{1}{5(x-1)}
$$
satisfies $\lim_{x \to 1} g(x) = -\frac{2}{5}$. Now, Taylor expanding $g$ around the point $1$ (you can expect $g$ to have a Taylor expansion : that's not a worry)
$$
\frac{1}{x^5-1} - \frac{1}{5(x-1)} \approx - \frac{2}{5} + O((x-1))
$$
near the point $x=1$ (leaving out everything except the constant term in the Taylor expansion of $g$).
Now, trying $x=1.01$ out gives $20.6$, which leads to $144 \times 2.06 = 2964.4$, which is also good enough : and avoids any kind of decimal by-hand division.
Note that we used the following heuristic in the second part : even when a function $h(x)$ is not differentiable at a point $x_0$ , it may happen that we can find $N$ so that $h(x)(x-x_0)^N$ is differentiable at $x_0$. In that case, we can still write down an asymptotic analysis for $h(x)$ near $x_0$. That's basically what we did here for the function $h(x) = \frac{1}{x^5-1}$ at $x_0=1$ with $N=1$, to find the nature of $h$ accurately.
That's an extension of the usual derivative approximation $h(x+a) \approx h(x) + ah'(x)+\ldots$ that we're considering. In complex analysis (where it is frequently used e.g. in the theory of generating functions), this kind of analysis appears when we are working with Laurent series , poles and meromorphic functions, for instance.
