i am learning maths so fast here in MSE, thank you guys so much for being here to help us!
so now, my next step towards proficiency: :).
i am trying to prove that $\binom{n}{k}\frac{1}{n^k}\leq\frac{1}{k!}$ for $n\geq1$ and for all $k\in\Bbb{N}$. My first problem is that i never tried to prove a statement with induction where i have two dependencies. here $n$ and $k$.
induction base: $n=1$ and $k=1$. $\binom{1}{1}\frac{1}{1^1}\leq\frac{1}{1!}$ which is okay.
inductive hypothesis: for $n\geq1$ and for all $k\in\Bbb{N},~~\binom{n}{k}\frac{1}{n^k}\leq\frac{1}{k!}$ 1)first induction step: $n\rightarrow n+1~~ \text{and}~~ k,\binom{n+1}{k}\frac{1}{(n+1)^{k}}\leq\frac{1}{(k)!}$ 2)second induction step: $n$ and $k\rightarrow k+1,~~\binom{n}{k+1}\frac{1}{(n)^{k+1}}\leq\frac{1}{(k+1)!}$
1)$\binom{n+1}{k}\frac{1}{(n+1)^{k}}=\Bigg(\binom{n}{k}+\binom{n}{k-1}\Bigg)\frac{1}{(n+1)^{k}} =\\ \binom{n}{k}\frac{1}{(n+1)^{k}} + \binom{n}{k-1}\frac{1}{(n+1)^{k}} \leq \frac{1}{(k)!}$
2) Help.
I am having hard time to prove further in calculations, can you pls show me further? is 1) right? i am not sure