If $F_n$ is the Fibonacci sequence, show that $F_n < \left(\frac 74\right)^n$ for $n\geq 1.$ Recall that the Fibonacci sequence is defined by $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n−1} + F_{n−2}$ for $n ≥ 2$. Prove that:
$$\forall \,\, n ≥ 1 ,\,\, F_n < \left(\frac 74\right)^n$$ 
In this question I understand how to do the basis step.
In the induction step I know that you have to assume that n=k but I am having trouble figuring out on how to do that. Could someone please explain how to do this question.

The proposition that you're trying to prove is that $F_n<(\frac{7}{4})^n$
For $n = 0$, this is trivial; $0 < (\frac{7}{4})^0$
For $n = 1$, we have $1 < (\frac{7}{4})^1$
For your induction step, you assume that for all k < n, $F_k<(\frac{7}{4})^k$
So $F_{n-2}<(\frac{7}{4})^{n-2}$ and $F_{n-1}<(\frac{7}{4})^{n-1}$
$F_{n} = $
$F_{n-2}+ F_{n-1}$ <
$(\frac{7}{4})^{n-2} + (\frac{7}{4})^{n-1}$ =
$(\frac{7}{4})^{n-2} + \frac 7 4 (\frac{7}{4})^{n-2} = $
$\frac 4 4(\frac{7}{4})^{n-2} + \frac 7 4 (\frac{7}{4})^{n-2} $=
$\frac {11} 4 (\frac{7}{4})^{n-2}$=
$\frac {44} {16} (\frac{7}{4})^{n-2}$<
$\frac {49} {16} (\frac{7}{4})^{n-2}$=
$ (\frac{7}{4})^{n}$
