Proof by Induction with n as Denominator and Exponent

Question

I need to prove the following inequality by induction, but I get stuck on every attempt I make. I simply keep getting stuck because $n$ is both a denominator and an exponent, can anyone help? $$\forall n \in \Bbb N , n \ge 2 : n<(2-\frac{1}{n})^n$$

Answer 1

Notice that $n^2 > n + 1$ for $n \geq 2$. This would imply $2n^2 - n > n^2 + 1$, or what is the same $2n^2 + 2n - n > \left( n + 1 \right)^2$. Hence, $2n - \dfrac{n}{n + 1} > n + 1$.

Also, $\left( 2 - \dfrac{1}{n + 1} \right)^n > \left( 2 - \dfrac{1}{n} \right)^n$.

Now, the statement is true for $n = 2$. Assume the induction hypothesis and we proceed for the case $n + 1$.

\begin{align*} \left( 2 - \dfrac{1}{n + 1} \right)^{n + 1} &= \left( 2 - \dfrac{1}{n + 1} \right)^n \left( 2 - \dfrac{1}{n + 1} \right) \\ &> \left( 2 - \dfrac{1}{n} \right)^n \left( 2 - \dfrac{1}{n + 1} \right) \\ &> n \left( 2 - \dfrac{1}{n + 1} \right) \\ &> n + 1 \end{align*}