Math contest: Find number of roots of $F(x)=\frac{n}{2}$ involving a strange integral. 
Edit summary: A good answer appeared. CW full answer added, based on given answers. Removing my ugly-looking attempts, as they still remain in the rev. history.

Here's a final-round calculus contest problem in our district in 2009. Test paper isn't available online. I originally proved $F(n/2)<n/2$ but failed to prove $F(n)>n/2$ and the monotonicity of $F(n)$.

Define
      $$F(x)=\int\limits_0^x \mathrm e^{-t} \left(1+t+\frac{t^2}{2!}+\cdots+\frac{t^n}{n!}\right) \mathrm d t$$
    where $n>1, n \in \mathbb N^+$; and an equation: $\displaystyle F(x)=\frac{n}{2}$.
Find the number of roots of this equation within the interval $\displaystyle I=\left(\frac{n}{2},n\right)$.

Tedious Previous Attempts Removed.

Proving $\displaystyle F(n)>\frac{n}{2}$:
Looks like you've made the life far more complicated than needed. For $t>0$, write
$$
(1-\frac tn)e^t=\sum_{k\ge 0}\frac 1{k!}(1-\frac kn)t^k<\sum_{k=0}^n \frac{t^k}{k!}
$$
(drop the negative coefficients and increase the rest).
Rewrite this as $$e^{-t}\sum_{k=0}^n \frac{t^k}{k!}> 1-\frac tn.$$ Now, I hope even a dumb calculus student can integrate the right hand side from $0$ to $n$.
