Prove that: $(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + (S_n)^2/2! + ... + (S_n)^n/n!$ I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next.
Question
If $a_1, a_2, ...,a_n$ are all positive and $S_n=a_1+a_2+...+a_n$ then:
$(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}$
My Attempt
Proof by induction:
I had first shown that it was true for $n=1$ and $n=2$.
Now suppose that it is true for n. Then it must also be true for $n+1$
$(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1}) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!} + \dfrac{(S_{n})^{n+1}}{(n+1)!}$
Define: $(1+a_1)(1+a_2)...(1+a_n)=x$ and $1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}=y$
Then: $x+xa_{n+1} \le y+\dfrac{(S_{n})^{n+1}}{(n+1)!}$
By the induction assumption, we know that $x \le y$.
What I can't figure out
From this point, what I think the next natural step would be is to show that
$xa_{n+1} \le \dfrac{(S_{n})^{n+1}}{(n+1)!}$
I know this rearranges to:
$xa_{n+1} \le \dfrac{(S_{n})^{n}}{n!} \times \dfrac{S_n}{(n+1)} $
And feel there may be something I can do here, but haven't been able to figure anything out.

Continue the proof using MI. Assume P(n) is true.
Let $y=a_{n+1}, S=S_n$
Let $f(y)=1+(S+y)+\frac{(S+y)^2}{2!}+...+\frac{(S+y)^{(n+1)}}{(n+1)!}-(1+a_1)(1+a_2)...(1+a_n)(1+y) \quad y>0$
$f'(y)=1+(S+y)+\frac{(S+y)^2}{2!}+...+\frac{(S+y)^{n}}{n!}-(1+a_1)(1+a_2)...(1+a_n)$
$ > 1+S+\frac{S^2}{2!}+...+\frac{S^{n}}{n!}-(1+a_1)(1+a_2)...(1+a_n) \geq 0 $
$f(0) = 1+S+\frac{S^2}{2!}+...+\frac{S^{(k+1)}}{(k+1)!}-(1+a_1)(1+a_2)...(1+a_n)$
$> 1+S+\frac{S^2}{2!}+...+\frac{S^{n}}{n!}-(1+x_1)(1+x_2)...(1+x_n) \geq 0$
Since f is differentiable w.r.t y, thus $f(y) \geq f(0) \geq 0$ for $y>0$
Rewrite $(S+y)$ as $S_{n+1}$ and $y$ as $a_{n+1}$
$1+S_{n+1}+\frac{S_{n+1}^2}{2!}+...+\frac{S_{n+1}^{(n+1)}}{(n+1)!} \geq (1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1})$
Then P(n+1) is also true.
