$$1.01^{5}=1.0510100501$$
Explanation:
By the binomial theorem, in general we have:
$$(a+b)^{n}=\displaystyle\sum_{k=0}^{n}\begin{pmatrix} n \\ k \end{pmatrix} a^{n-k}b^{k}$$
where $$\begin{pmatrix} n \\ k \end{pmatrix}=\dfrac{n!}{(n-k)!k!}$$
The binomial coefficient $$\begin{pmatrix} n \\ k \end{pmatrix}$$ occurs as the $$(k+1)$$ th term in the $$(n+1)$$th row of Pascal's triangle. in our example, we are interested in the sixth row, which consists of the terms:
$$1,5,10,5,1$$
So we find:
$$1.01^{5}=(1+0.01)^{5}$$
$$=1+5(0.01)+10(0.01)^{2}+10(0.01)^{3}+5(0.01)^{4}+(0.01)^{5}$$
Footnote
Note that for small powers $$n$$, this gives you a way to find a row of binomial coefficients on a calculator.
Just calculate $$1.01^{n}$$ and read the digits in pairs. For example:
$$1.01^{4}=1.04060401$$