I want to show that $(n+1)^{n-1}<n^n$ using the binomial theorem.
I have done the following:\begin{align*}\left (n+1\right )^{n-1}&=\sum_{k=0}^{n-1}\binom{n-1}{k}n^{n-1-k}1^k=\sum_{k=0}^{n-1}\binom{n-1}{k}n^{n-1-k}\\ & <\sum_{k=0}^{n-1}\binom{n-1}{k}n^{n-1}=n^{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k} \\ & =n^{n-1}\cdot 2^{n-1}<n^{n-1}\cdot n^{n-1}=n^{2n-2}\end{align*} This is not correct.
What could I change to get the correct result?
This is equivalent to showing that $\left(1+\frac1n\right)^{n-1}<n,$ that is, that
$$1+n\left(\frac1n\right)+\frac{n(n-1)}{2}\left(\frac1n\right)^2+\cdots+\left(\frac1n\right)^{n-1}<n.$$
But this is true since the expansion can be rearranged thus
$$ 1+1+\frac12\left(1-\frac1n\right)+\frac{1}{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots+\frac{1}{n^{n-1}},$$ from which it is obvious that the inequality follows, since the expansion is less than $$\underbrace{1+1+1+\cdots+1+1}.$$