discrete mathematics - Help finishing proof via induction for a summation

So I have to prove the following equation using induction for n >= 2:
$$\sum\limits_{i=1}^n 4/5^i < 1$$
However the question asks me to prove something stronger such as this:

$$\sum\limits_{i=1}^n 4/5^i <= 1 - \frac{1}{5^n}$$
first to imply the first equation is true.
So far I have the following:

Base Case:
Let n = 2

$$\sum\limits_{i=1}^2 4/5^i = \frac{4}{5} + \frac{4}{25} = \frac {24}{25}$$
then I also applied it to
$$1 - \frac{1}{5^n} \rightarrow 1 - \frac{1}{5^2} = \frac{24}{25}$$
Therefore I can make the following assumptions yes?

Inductive Hypothesis
for all 2 <= n <= k it is
$$\sum\limits_{i=1}^n 4/5^i = 4\frac{\frac{1}{5^n} - 1}{\frac{1}{5} - 1} = 1 - \frac{1}{5^n} < 1$$
Inductive Step
Hopefully I'm ok up to here, I'll show what I have so far for this step.
$$\sum\limits_{i=1}^{k+1} 4/5^i = \frac{\frac{1}{5^{k+1}} - 1}{\frac{1}{5} - 1} = 4\frac{(\frac{1}{5^k}-1) * \frac{1}{5} - \frac{4}{5}}{\frac{1}{5} -1}$$
$$= \frac{1}{5} * 4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1} - 4\frac{\frac{4}{5}}{\frac{1}{5} - 1}$$
so here I have:
$$4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1}$$
which I know is:
$$= \sum\limits_{i=1}^k 4/5^i$$
which is my inductive hypothesis, I am unsure of how to finish my proof from here... any help correcting or finishing the proof is very much appreciated

Answer

Hint:
$$\sum_{i=1}^{k+1}4/5^i=\sum_{i=1}^k 4/5^i+4/5^{k+1}$$

Use the induction hypothesis on the sum from $1$ to $k$ and simplify.