proving that $(n-1)^n>n^{n-1}$ [duplicate]

Solution 1:

Solution using induction:

Base case - $n=4$, so $(4-1)^4>4^{4-1}$ which is true.

Inductive step - assume that the statement holds for some k and show that it holds for k+1, hence we want to show that $k^{k+1}>(k+1)^k$ for $k \ge 3$.

Our assumption is $(k-1)^k>k^{k-1}$ or alternatively $\displaystyle k-1>\left(1+\frac{1}{k-1}\right)^{k-1}$.

Dividing both sides by $k^k$ we get that we need to show that $\displaystyle k>\left(1+\frac{1}{k}\right)^k$ for $k \ge 3$.

It is obvious that $\displaystyle 1+\frac{1}{k}<1+\frac{1}{k-1} \rightarrow \left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{k-1}\right)^k$, so it is sufficient to show that $\displaystyle k>\left(1+\frac{1}{k-1}\right)^k$.

We can write: $$\displaystyle \left(1+\frac{1}{k-1}\right)^k=\left(1+\frac{1}{k-1}\right)\cdot{\left(1+\frac{1}{k-1}\right)^{k-1}}\underbrace{<}_{using \ assumption}\left(1+\frac{1}{k-1}\right)(k-1)=k$$

Solution 2:

take the logarithm of both sides and you will get $\frac{\ln(n-1)}{n-1}>\frac{\ln(n)}{n}$ to prove this consider the function $f(x)=\frac{\ln(x)}{x}$