I have some issues proving that ${n \choose s} \le (\frac{ne}{s})^s$ for $n,s \in N$ (Newton binomial). I've tried to do it by mathematics induction (although if there's another way, I'm open for it), assuming that $s \le n$, but I was told that this is wrong because the s is restricted by the n value, so I should either induce using n, or use something called a "generalized formula" and take all the cases of the $s$ values into consideration. And this is where my mind just faded out, I simply cannot move forward.
Could anyone explain what am I doing wrong here, or provide me with the correct solution so that I know where did my path differ? Below is what I have currently with some explanation for every step.
I am assuming that $s \le n$, and I'm proving the thesis for $s=1$: $${n \choose 1} \le (\frac{ne}{1})^1 <=> \frac{n!}{1!(n-1)!} \le ne <=> n \le ne <=> 1 \le e$$
Then, for some number $s \ge 1$, I'm proving the thesis for $s+1$: $$\frac{n!}{(s+1)!(n-s-1)!} \le (\frac{ne}{s+1})^{s+1}$$ $$\frac{n!}{s!(n-s)!\frac{s+1}{n-s}} \le (\frac{ne}{s+1})^{s+1}$$
What I've got is the Newton Symbol for n and s $${n \choose s} \le (\frac{ne}{s+1})^{s+1}\frac{s+1}{n-s}$$
Then I'm trying to transform it so that I am getting the exact expression from inequality from the thesis I'm proving: $${n \choose s} \le \frac{(ne)^s ne}{(s+1)^s}*\frac{1}{n-s}$$ $$...$$ $${n \choose s} \le (\frac{ne}{s})^s * \frac{s^s}{(s+1)^s} * \frac{ne}{n-s}$$
We already know that $(\frac{ne}{s})^s \ge {n \choose s} <=> \frac{(\frac{ne}{s})^s}{n \choose s} \ge 1$ and we can transform it to $$\frac{s^s}{(s+1)^s} * \frac{ne}{n-s} * \frac{(\frac{ne}{s})^s}{n \choose s} \ge 1$$.
So we need to prove only the part of this expression: $\frac{s^s}{(s+1)^s} * \frac{ne}{n-s} \ge 1$
To do it I use the expression true for every natural number $s$: $(\frac{s+1}{s})^s \le e$, that can be further transformed to $e * \frac{s^s}{(s+1)^s} * \frac{n}{n-s} \ge 1$
Which is true when $\frac{n}{n-s} \ge 1 <=> n \ge n-s$, and this is correct with our assumption that $s \le n$, which is ending the proof.
That is all. I apologize for my English, as I do not use the math nomenclature too often.