Proving logarithmic inequality

Question

I was looking through my old notebook from uni and I stumbled upon this exercise. Sadly I couldn’t solve it then I cannot do it now. What would be best approach with this question? For $a>2$ and $n>0$, prove using induction that $\log_a(1+n)<n$. I’m not sure why $a$ has to be larger than 2. I’ve plotted both sides of inequality and graphically makes sense. I’ve tried to raise both sides to the power of $a$, but that’s as far I could get with it.

Answer 1

$$a>2 \implies \log_a2<1$$ This means that for $n=1$, $\log_a(1+1)<2$.

For $n=k$, $$\log_a(1+k)<k \implies a^k>1+k \implies a^{k+1}>a(1+k)>2(1+k)>2+k $$ $$\implies \log_a(2+k)<k+1$$ Hence, proved by induction.