Step 1: Simplify with a substitution before differentiating.
Let $x = r/a$, a dimensionless distance. Then the radial probability distribution becomes:
\[ P(r) \propto r^4 e^{-r/a} = a^4x^4 \cdot e^{-x} \]
Since $a^4$ is just a constant multiplier, maximizing $P$ over $r$ is the same as maximizing $g(x) = x^4e^{-x}$ over $x$.
Step 2: Use logarithmic differentiation instead of the product rule.
Take the natural log of $g(x)$ (valid for $x>0$, where $g>0$):
\[ \ln g(x) = 4\ln x - x \]
At a maximum, the derivative of $\ln g(x)$ is zero, since $\ln$ is an increasing function and $g$, $\ln g$ peak at the same $x$:
\[ \dfrac{d}{dx}\ln g(x) = \dfrac{4}{x} - 1 = 0 \]
Step 3: Solve for $x$ and convert back to $r$.
\[ \dfrac{4}{x} = 1 \implies x = 4 \]
Since $x=r/a$, this gives $r = 4a$. To confirm this is a maximum, check that $\ln g(x)$ is increasing for $x<4$ (since $4/x>1$ there) and decreasing for $x>4$ (since $4/x<1$ there), which is exactly the shape of a single peak.
Final Answer:
The radial probability distribution peaks at $r=4a$, so comparing with $r=n_0a$ gives $n_0=4$.
\[ \boxed{n_0 = 4} \]