Inequalities with Exponential Functions

First, find the range of natural numbers $n$ satisfying $5^{n-1} < 3^{n+1}$: $\frac{5^n}{5} < 3 \cdot 3^n \implies \left(\frac{5}{3}\right)^n < 15$

Testing natural numbers $n$:

• $n=1: 5/3 < 15$ (True)
• $n=2: 25/9 \approx 2.78 < 15$ (True)
• $n=3: 125/27 \approx 4.63 < 15$ (True)
• $n=4: 625/81 \approx 7.72 < 15$ (True)
• $n=5: 3125/243 \approx 12.86 < 15$ (True)
• $n=6: 15625/729 \approx 21.43 < 15$ (False)

So $n \in \{1, 2, 3, 4, 5\}$.

We need $3^{n+1} < 2^{n+m}$ to hold for all $n \in \{1, 2, 3, 4, 5\}$. Since $3^{n+1} / 2^n$ increases as $n$ increases, the strongest condition occurs at $n = 5$: $3^{5+1} < 2^{5+m} \implies 3^6 < 2^{5+m} \implies 729 < 2^{5+m}$

Since $2^9 = 512$ and $2^{10} = 1024$, we need: $5 + m \ge 10 \implies m \ge 5$

Thus, the least integer value of $m$ is $5$.