We must show that $(n+1)^3$ is greater than $3n$ for all $n$ that is a natural number less than or equal to 4. The natural numbers less than or equal to 4 include 1, 2, 3, and 4. Hence, proving the above by exhaustion will involve showing that it is true for $n$ set equal to 1, 2, 3 and 4:

- For $n=1$, $(n+1)^3=(1+1)^3=2^3=8$ and $3^n=3^1=3$. Since, $8\geq 3$, the above is true when $n=1$.
- For $n=2$, $(n+1)^3=(2+1)^3=3^3=27$ and $3^n=3^2=9$. Since, $27\geq 9$, the above is true when $n=2$.
- For $n=3$, $(n+1)^3=(3+1)^3=4^3=64$ and $3^n=3^3=27$. Since, $64\geq 27$, the above is true when $n=3$.
- For $n=4$, $(n+1)^3=(4+1)^3=5^3=125$ and $3^n=3^4=81$. Since, $125\geq 81$, the above is true when $n=4$ and thus concluding the proof by exhaustion.

Note that $n=5$ is the first instance when the statement breaks down.