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\textbf{Solution 2: }  \textbf{Proof:}  \\*  Base Case: For $n=3$, $2^3+3^3+4^3 \\$2^3+3^3+4^3  \leq 5^3 = 99 5^3$ \\$99  \leq 125$. The \\The  inequality holds true for $n=3$ \\  Assume: $2^k+3^k+4^k\leq5^k$. We must show that $k+1$, $2^{k+1}+3^{k+1}+4^{k+1}\leq5^{k+1}$