Notice first that
$$3^{2} + 4^{2} = 5^{2}, \quad \text{and} \quad 3^{3} + 4^{3} + 5^{3} = 6^{3}.$$
However
$$3^{4} + 4^{4} + 5^{4} + 6^{4} = 2258 < 2401 = 7^{4}$$
Assume that
$$3^{n} + 4^{n} + \cdots + (n+2)^{n} < (n+3)^{n}$$
for some $n \geq 4$. We have
$$\begin{aligned} 3^{n+1} + 4^{n+1} + & \cdots + (n+2)^{n+1} + (n+3)^{n+1} \\ & < (n+2)\left(3^{n} + 4^{n} + \cdots + (n+2)^{n}\right) + (n+3)^{n+1} \\ & < (n+2)(n+3)^{n} + (n+3)^{n+1} < 2(n+3)^{n+1} \end{aligned}$$
For $n=4$, we have $(n+2)(n+3)^{n} + (n+3)^{n+1} = 31213 < 32768 = (n+4)^{n+1}$. For $n \geq 5$, we have
$$\begin{aligned} \left(\frac{n+4}{n+3}\right)^{n+1} &= \left(1 + \frac{1}{n+3}\right)^{n+1} > 1 + \frac{n+1}{n+3} + \frac{n(n+1)}{2(n+3)^{2}} \\ & > 2 + \frac{(n-5)(n+2) + 2}{2(n+3)^{2}} > 2 \end{aligned}$$
Therefore, $3^{n+1} + 4^{n+1} + \cdots + (n+2)^{n+1} + (n+3)^{n+1} < (n+4)^{n+1}$.
Hence, we have proved by induction that $3^{n} + 4^{n} + \cdots + (n+2)^{n} < (n+3)^{n}$ for all $n \geq 4$.