Proof $(1+i)^t < 1+it$ if $0 < t < 1$

Question

I am trying to prove that compound interests grow slower than simple interest in the first year.

I attempted to do this with the following approach.

Let $$f(t) = (1+i)^t$$

then $$f'(t) = \ln({1+i})(1+i)^t$$

while if $$g(t) = 1+it$$

then $$g'(t) = i$$ .

I was able to show that $f(0) = g(0)$ and since we are implicitly assuming that interest $i$ is positive, $i> \ln{1+i}$ shows that simple interest, indeed, starts out growing faster than compound interest. However, I am having trouble that this ends at $t=1$.

I'm not sure if this is even an efficient approach, but I would really like it if

a), If someone could help me out with the second part.

b), Give an alternative solution that would be easier.

Answer 1

Just remark that, by a Taylor expansion, $$ (1+i)^t = 1+it+\frac{\sigma(\sigma-1)}{2}i^2 $$ for some $\sigma$ between $0$ and $t$. Therefore $\sigma-1<0$, and you conclude.

Answer 2

Derive on $i$:

$$t(1+i)^{t-1}<t$$ because the exponent is negative.