Show by induction that $n! > n \cdot F_n$ for all $n > 3$ Show by induction that $n! > n \cdot F_n$ for all $n> 3$, where $F_n$ is the $n$th fibonacci number. 
For the basis step, I put $n=4$ and got:
\begin{align*}
&4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\
&4 \cdot F_4 = 4 \cdot 3 = 12
\end{align*}
So the statement holds for $n=4$. For the inductive step, we assume the statement holds for a $k > 3$, and checks for $n = k+1$. We have:
\begin{align*}
&(k+1)! = (k+1)\cdot k \cdot (k-1) \cdot \; \dotsc \; \cdot 1 = (k+1 ) \cdot k!  \\
&(k+1) \cdot F_{k+1} = (k+1)(F_k + F_{k-1}) = k \cdot F_k +  k\cdot F_{k-1} + F_{k} + F_{k-1}
\end{align*}
By the induction assumption, we know that
\begin{align*}
k\cdot F_k + (k-1) \cdot F_{k-1} < k! + (k-1)! \leq k! + k! < 2k! <(k+1)k!,
\end{align*}
But we see that 
\begin{align*}
k\cdot F_k + (k-1) \cdot F_{k-1}  < (k+1) \cdot F_{k+1},
\end{align*}
so we can't really say if $(k+1) \cdot F_{k+1}$ is less than $(k+1)!$ or not. 
Does this mean that the statement is not true, or is there just a better way of proving it?

How about the following way?
It is sufficient to prove that $(n-1)!\gt F_n$ for $n\gt 3$.
Inductive step : we assume that $(k-1)!\gt F_k$.
Then, multiplying the both sides by $k$ gives
$$k!\gt kF_k\gt F_k+F_k\gt F_k+F_{k-1}=F_{k+1}$$
