How many ways can a natural number n be expressed as a sum of one or more positive integers, taking order into account?

One way to see why the answer is $2^{n-1}$ directly is to write $n$ as a sum of $1$s:

$$ n = \underbrace{1 + 1 + 1 + ... + 1}_{n \textrm{ times}}. $$

Of course, this expresses $n$ as a sum of $1$s, but we want all expressions of $n$ as a sum of (ordered) positive numbers.  To do so, for each '$+$' in the expression above, we can choose whether or not to split the sum there to form a new summand. For example, suppose $n = 8$.  Then we start with

$$ 8 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1. $$

Now suppose we choose to split the sum at the 2nd, 3rd and 5th plus signs, highlighted below.

$$ 8 = 1 + 1 \color{red}{+} 1 \color{red}{+} 1 + 1 \color{red}{+} 1 + 1 + 1. $$

We now add up the $1$'s between the chosen plusses to form the summands,

$$ 8 = (1+1) \color{red}{+} 1 \color{red}{+} (1+1) \color{red}{+} (1+1+1) = 2 \color{red}{+} 1 \color{red}{+} 2 \color{red}{+} 3,$$

giving an expression of $8$ as a sum of positive integers.

In general, this clearly bijects to all expressions of $n$ as an ordered sum of positive integers. Since there are $n-1$ plus signs between the $n$ $1$s, there are $2^{n-1}$ ways of choosing where to split the sum, and hence $2^{n-1}$ possible sums.

What you are looking for is compositions of $n$

Take a string of $1's,\;$ e.g.$\; \large 1{\boxed.} 1\boxed. 1\boxed. 1$

In the $\;(n-1)\;$ boxes, either put a $+\;$ or a comma.

$1,\;\;1+1+1$ e.g. would represent $1+3$

Since you have $2$ choices for each box, # of compositions = $2^{n-1}$