I am asked to show that $\forall\space x>0, \forall\space n\in\mathbb{N}$ and for every $0\le k\le n$: $$(1+x)^n > \frac{n(n-1)...(n-k+1)}{k!}x^k$$
Using Bernoulli only gives me $(1+x)^n>1+nx$ which I am not being able to use. If I start from the RHS, I have: $$\frac{n(n-1)...(n-k+1)}{k!}x^k\le \frac{n^kx^k}{k!}$$
However, I already have a counter example to the following statement: $$\frac{n^kx^k}{k!}\le (1+x)^n$$
Which is to let $n=2, k=2$ and $x\ge\frac{5}{2}$
This means that my approximation is already too big. Any tips on how to proceed?
If $x>0$ then $$(1+x)^n =\sum_{i=0}^n \binom{n}{i} x^i > \binom{n}{k} x^k \text{ for }0\leq k \leq n.$$