TITLE: prove $\binom{n}{k}\frac{1}{n^k}\leq\frac{1}{k!}$
QUESTION [14 upvotes]: 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

REPLY [7 votes]: In general, for inductions involving two independent natural numbers (here, $n$ and $k$, with $n, k \ge 1$), the strategy is as follows: 

You first prove the basis case for $n=k=1$, which you have done. 
Then you state an inductive hypothesis that the property holds for $n = a$ and $k = b$. 
Then you need two inductive steps, and for each you assume the inductive hypothesis to be true, i.e. you'll want to use the inductive hypothesis to

First, prove the hypothesis also holds for $n=a+1$ and $k = b$; and 
Then, prove that the hypothesis also holds for $ n=a, k = b+1$.