Obtain the equation problem If you have an amount of 1€, and you receive 100% interest at the end of the year, your amount will be 2€ after one year. If you instead receive 50% interest every half year, you will have $1.5^2=2.25€$. If you receive 25% interest every quarter, you will have $1.25^4=2.44€$.
What amount will you have if you receive the interest $n$ times per year?
Can someone explain me how he gets to the conclusion that
$(1+1/n)^n$ is the solution?
I understand that the first year the result will be: $1+\frac{1}{2}=\frac{3}{2}$
I understand that the second year will be: $(1+\frac{1}{2})+(\frac{3}{2}\cdot\frac{1}{2})$
It just doesn’t get into my head where the power enters:
I understand that factorizing the formula: $(1+\frac{1}{2})\cdot(1+\frac{1}{2})$
you’ll get somewhere but I don’t understand
 A: Imagine you have your amount of $1€$ and you'll get $50\%$ interest two times a year, so after every 6 month. After the first half you will have $1€ \cdot 1.5 = 1.5€$, where we use the factor $1.5$ because you increase the amount of you money by $50\%$.
So far so good. Now after the second half of the year you'll get again $50\%$ interest on whatever you have in you bank account, in this case it's $1.5€$ from the first half. So we can do the calculation again: $1.5€ \cdot 1.5 = 2.25€$. Notice that we can rewrite this to
$$(\underbrace{\underbrace{(1€ \cdot 1.5)}_{\text{first half}} \cdot 1.5)}_{\text{second half}} = 1€ \cdot 1.5^2 = 2.25€$$
This can be applied to $n$ amounts per year, where your interest rate is $1/n$ and the factor to multiply your money is $1+1/n$:
$$(\underbrace{\dots(\underbrace{\underbrace{(1€ \cdot (1+1/n))}_{\text{first nth of the year}} \cdot (1+1/n)}_{\text{second nth of the year}}) \dots (1+1/n))}_{\text{last nth of the year}} = 1€ \cdot (1+1/n)^n$$
I hope this somewhat clarifies where the solution $(1+1/n)^n$ and the powers come from.
Fun Fact:
For $n \to \infty$ the factor will be Euler's number.
$$\underset{n \to \infty}{\lim}(1+1/n)^n = e \approx 2.7182$$
