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!

Thanks to all (including those who helped in Pure algebra: Show that this expression is positive) we have shown the result. I will summarize it for those who can be interested.
Note that $B(g,n)$ simplifies to
$B(g,n)=\sum _{k=1}^n \frac{n!}{k!(n-k)!}g^{n-k}(1-g)^{k-1}(1-(1-f(g))^k)=\frac{-1+(1-f(g)+gf(g))^n}{-1+g}$.
Hence
$\frac{dB(g,n)}{dg}=\frac{(g-1)n(g-(g-1)(1-f(g)))^{n-1}(f(g)+(g-1)f'(g))-((g-(g-1)(1-f(g)))^n-1)}{(g-1)^2}$.
Because $\frac{dB(g,n)}{dg}$ is increasing in $f'(g)$, and we have assumed that $f'(g)\geq0$, we can can now set $f'(g)=0$. This simplifies the previous derivative to
$\frac{dB(g,n)}{dg}=\frac{1-(1-g)nf(g)(1-(1-g)f(g))^{n-1}-(1-(1-g)f(g))^n}{(g-1)^2}.$
We want to show that this derivative is positive. To do so, set $t=1-(1-g)f(g)$ in the previous expression and eliminate the (positive) denominator. Now we need to show that
$f_n(t):= 1-n(1-t)t^{n-1}+t^n=1-nt^{n-1}+(n+1)t^n>0$.
Note that
$f'_n(t)=-n(n-1)t^{n-2}+n(n+1)t^{n-1}$,
where $t_0=0$ and $t_n=\frac{n-1}{n+1}$ make $f'_n(t)=0$. Since $f''_n(t_n)>0$, we conclude that $t_n$ is a minimum.
Now remains to show that $f_t(t_n)>0$.
$f_t(t_n)=1-n(\frac{n-1}{n+1})^{n-1}+(n+1)(\frac{n-1}{n+1})^{n}=1-(\frac{n-1}{n+1})^{n-1}>0$.
Thanks to everyone who contributed!
