As Pioncarė noted that Mathematics is the art of giving new name to same things. So, if $p^n$ can be expressed as the sum of ramps, let us call it ramped. The most obvious thing to do is get your hands dirty. Let us try $p=2$ and $n=1.$ Let there be positive integers such that $ a \mid b$ and $b \mid c,$ and
$$ 2 = a+b+c $$
Since $a,b,c$ are positive integers, $a+b+c \ge 3.$ Therefore, no such ramp exists for this case. Wait! That does not mean that no other powers of $2$ can have a ramp. I mean an off-ramp is possible. Indeed, there is:
$$ 2^2 = 1+ 1+ 2 $$
Ah, and there it is. Multiply the previous equation by $2^{n-2}$ to get:
$$ 2^n = 2^{n-2} + 2^{n-2} + 2^{n-1} \ \text{for all} \ n \ge 2. $$
So, $2^n$ can be ramped except for $n=1.$ Similarly,
$$ 3 = 1+1+1 $$
$$ \implies 3^n = 3^{n-1} + 3^{n-1}+ 3^{n-1} $$
So, all the powers of $3$ can be ramped. If we think about $5,$ then
$$ 5 = 1 + 2+ 2 $$
$$ \implies 5^n = 5^{n-1} + 2 \cdot 5^{n-1} + 2 \cdot 5^{n-1} $$
So, all the powers of $5$ can be ramped. Similarly,
$$ 7 = 1 + 2 + 4 $$
$$ \implies 7^n = 7^{n-1} + 2 \cdot 7^{n-1} + 4 \cdot 7^{n-1}. $$
So, all the powers of $7$ can be ramped. Have we discovered a trick here? Indeed, we have. Let $p$ be a prime number greater than $7,$ then
$$ p = 1+ 2 + (p-3) $$
Since all prime numbers greater than $7$ are odd, therefore $(p-3)$ is even and $ p-3 > 4. $ Therefore, notice that $1,2,(p-3)$ is a ramp.
$$ p^n = p^{n-1} + 2 \cdot p^{n-1} + (p-3) \cdot p^{n-1}. $$
Therefore, all prime numbers $p$ for all positive integers $n$ can be ramped except for $p=2$ and $n=1.$
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