Define "f"(n) = sum_(k=0)^n s^k/(k!).
"f"'(n)= sum_(k=0)^n ks^(k-1)/(k!),
"f"'(n) = sum_(k=0)^(n-1) s^k/(k!),
"f"'(n) = "f"(n-1).
"f"(n) is increasing for n in ZZ^+ if "f"'(n)>0. "f"'(n)>0 if "f"(n-1)>0. "f"(0)=1. Inductively, "f"(n) is increasing for n in ZZ^+ and therefore "f"(n)>0.
I'm going to prove a useful inequality.
Consider "g"(x) = x-ln(1+x). Then "g"'(x) = 1 - 1/(1+x). Clearly 1/(1+x) < 1 for x>0, so we conclude that "g"(x) is an increasing function. As "g"(0) = 0, "g"(x)>0 for x>0. Then,
ln(1+x)<x
As x>0 and a_k>0, let x=a_k. We have,
ln(1+a_k)<a_k.
Summing each side,
sum_(k=1)^(n) ln(1+a_k) < sum_(k=1)^n a_k.
We can substitute our definition of s.
sum_(k=1)^n ln(1+a_k) < s.
By using ln(a)+ln(b)=ln(ab) and writing s=ln(e^s) we conclude,
ln(prod_(k=1)^n (1+a_k)) < ln(e^s).
By taking an inverse logarithm,
prod_(k=1)^n (1+a_k) < e^s.
Define "h"(n) = sum_(n+1)^(\infty) s^k/(k!).
Then,
"h"'(n) = sum_(k=n+1)^(\infty) k * s^(k-1)/(k!),
"h"'(n) = sum_(k=n)^(\infty) s^k/(k!).
"h"'(n) = "h"(n-1).
Then "h"(n)>0 if "h"(n-1)>0. Inductively, as "h"(0)=e^s and e^s>0, "h"(n)>0.
So, we have that,
prod_(k=1)^n (1+a_k) < e^s,
prod_(k=1)^n (1+a_k) < sum_(k=1)^(n) s^k/(k!) - sum_(k=n+1)^(\infty) s^k/(k!),
prod_(k=1)^n (1+a_k) < sum_(k=1)^(n) s^k/(k!).
