Induction Proof $4^k < 5^k$

Question

So I need to use induction to prove that $4^n<5^n$ for all natural numbers $n > 0$. I have proved the base case to be true. For the induction hypothesis, we assume that when $n=k$, then $4^k < 5^k$. However, I am stuck in the inductive step when I try to show that it needs to be true for $n+1$. Therefore, we need to show that $4^{k+1} < 5^{k+1}$. However I get stuck on the first step trying to prove it as in

1. $4^{k+1} = 4 \cdot 4^k$, then by substituting $5^k$ instead of $4^k$ from the inductive hypothesis and I get stuck right here and I do not know how to proceed.

Any help would be appreciated.

Answer 1

When $n = 1$, clearly $4 < 5$. Suppose inductively that $4^k < 5^k$. Then \begin{align*} 4^{k+1} = 4^k \cdot 4 < 5^k \cdot 4 < 5^k \cdot 5 = 5^{k+1}. \end{align*}

Answer 2

Since $4^k < 5^k$ by the inductive hypothesis, we have $4^{k + 1} = 4 \cdot 4^k < 4 \cdot 5^k < 5 \cdot 5^k = 5^{k + 1}$. This completes the induction.