Show that the derivative of this function is positive 
Suppose that $n>1$, $g\in(0,1)$ and $f(g)\in(0,1)$. Suppose that 
  $\frac{df(g)}{dg}\geq0$.
Define $B(g,n)$ as:
$B(g,n)=\sum _{k=1}^n \frac{n!}{k!(n-k)!}g^{n-k}(1-g)^{k-1}(1-(1-f(g))^k)$
Show that:
$\frac{dB(g,n)}{dg}>0$.

Some of these assumptions can be relaxed (for example the $\frac{df(g)}{dg}\geq0$ I suspect), but I am not especially interested in that.
What I have done:


*

*I have shown that this is the case for $n=2$, $n=3$ and $n=4$, but did not find any pattern that helped me generalize to $n$.

*I took the derivative with Mathematica and obtained that little monster:
$
\frac{dB(g,n)}{dg}=-\frac{g^n \left(-n \left(\frac{(g-1) f(g)+1}{g}\right)^{n-1} \left(\frac{(g-1) f'(g)+f(g)}{g}-\frac{(g-1) f(g)+1}{g^2}\right)-n \left(\frac{1}{g}\right)^{n+1}\right)}{g-1}-\frac{n g^{n-1} \left(\left(\frac{1}{g}\right)^n-\left(\frac{(g-1) f(g)+1}{g}\right)^n\right)}{g-1}+\frac{g^n \left(\left(\frac{1}{g}\right)^n-\left(\frac{(g-1) f(g)+1}{g}\right)^n\right)}{(g-1)^2}$.

*(Here I might be wrong) I think that the problem can be slightly simplified by assuming that $\frac{df(g)}{dg}=0$. Because assuming $\frac{df(g)}{dg}>0$ only "helps us" in proving the that the derivative is positive, then it suffices to show that our desired result holds when $\frac{df(g)}{dg}=0$. 
Thanks in advance!

This is not a full solution, but some possibly helpful thoughts too long to fit a comment.
Notice that
$$B(g,n) = \sum _{k=1} ^n \frac {n!} {k!(n-k)!} g^{n-k} (1-g)^{k-1} - \sum _{k=1} ^n \frac {n!} {k!(n-k)!} g^{n-k} (1-g)^{k-1} (1-f(g))^k = \\
\frac 1 {1-g} \sum _{k=1} ^n \frac {n!} {k!(n-k)!} g^{n-k} (1-g)^k - \frac 1 {1-g} \sum _{k=1} ^n \frac {n!} {k!(n-k)!} g^{n-k} (1-g)^k (1-f(g))^k = \\
\frac 1 {1-g} \{ [g + (1-g)]^n - 1 \} - \frac 1 {1-g} \{ [g + (1-g)(1-f(g))]^n - 1 \} = \\
\frac 1 {g-1} \{ [g - (g-1)(1-f(g))]^n - 1 \} \ ,$$
whence
$$\frac {\Bbb d B(g,n)} {\Bbb d g} = \frac { \{n [g - (g-1)(1-f(g))]^{n-1} [1 - (1-f(g)) + (g-1) f'(g)] \} (g-1)} {(g-1)^2} - \frac {\{ [g - (g-1)(1-f(g))]^n - 1 \}} {(g-1)^2} = \\
\frac {\Bbb d B(g,n)} {\Bbb d g} = \frac { \{n [g - (g-1)(1-f(g))]^{n-1} [f(g) + (g-1) f'(g)] \} (g-1)} {(g-1)^2} - \frac {\{ [g - (g-1)(1-f(g))]^n - 1 \}} {(g-1)^2} \ .$$
Since the denominator is positive, you have then to show that
$$\{n [g - (g-1)(1-f(g))]^{n-1} [f(g) + (g-1) f'(g)] \} (g-1) - \{ [g - (g-1)(1-f(g))]^n - 1 \} \ge 0 \ .$$
Notice that
$$g - (g-1)(1-f(g)) = 1 + (g-1)f(g) < 1 \ ,$$
whence it follows that
$$- \{ [g - (g-1)(1-f(g))]^n - 1 \} \ge 0 \ .$$
and that
$$n [g - (g-1)(1-f(g))]^{n-1} \to 0 \ ,$$
so that $\lim \limits _{n \to \infty} \dfrac {\Bbb d B} {\Bbb d g} \ge 0$. Unfortunately, it is not clear whether this limit is uniform.
It also follows that a sufficient (but not necessary) condition to get the desired conclusion is $f + (g-1)f' \le 0$.
