Elementary proofs about $f(t)=\lim_{n\to\infty}(1+\frac tn)^n$ Please read carefully what I mean by elementary before marking the question as duplicate.
I'm starting calculus lessons in High School; that is: no series, no derivatives, no limits of functions, no continuity... I have just stated the axiom of supreme and defined sequence and limit of sequence. I have proved that a monotone, bounded sequence has limit. Just that.
Under so 'elementary' conditions, I'd like to prove:

*

*For each real $t$, the sequence $\left(1+\dfrac tn\right)^n$ converges. (I have problems for $t<0$). Then I'll define $e^t$ as the limit of this sequence.

*For every pair $t,s$, $e^{t+s}=e^te^s$.

Any ideas or bibliography?
 A: For convergence, I show that the sequence $\left(1+\dfrac tn\right)^n$ is increasing.
If $t>0$ and $n$ is big, this is seen as follows: $$\frac{\left(1+\dfrac t{n+1}\right)^{n+1}}{\left(1+\dfrac tn\right)^n} = \left(1+\dfrac t{n+1}\right) \left[ \frac{n(n+1+t)}{(n+t)(n+1)}\right]^n = \left(1+\dfrac t{n+1}\right)\left[1-\frac{t}{(n+t)(n+1)}\right]^n \geq^1$$
$$\geq^1\left(1+\dfrac t{n+1}\right)\left[1-\frac{nt}{(n+t)(n+1)}\right]\geq^2 \left(1+\dfrac t{n+1}\right)\left(\frac{1}{1+\frac{t}{n+1}}\right)=1$$Where inequality $\geq^1$ follows from $(1+x)^n \geq 1+nx$ for $x>-1$ (and this is where we are taking $n$ big enough so $1-\frac{t}{(n+t)(n+1)}>-1$) and inequality $\geq^2$ is algebraic manipulation: $$1-\frac{nt}{(n+t)(n+1)} = \frac{n^2+t+n}{n^2+nt+t+n} = \frac{1}{1+\frac{nt}{n^2+n+t}} = \frac{1}{1+\frac{t}{n+1+t/n}} \geq \frac{1}{1+\frac{t}{n+1}}$$
For $t<0$, the inequalities I have written can be seen to still hold if $n$ is big enough.
Also, the sequences are bounded. For $t<0$ this is easy because $\left(1+\dfrac tn\right)^n<1^n=1$. For $t>0$, fix an integer $m>\max \{t,2\}$, and write $$\left(1+\dfrac tn\right)^n \leq \left(1+\dfrac mn\right)^n = \sum_{k=0}^n\frac{n!}{(n-k)!}\frac{m^k}{n^k}\frac{1}{k!} \leq \sum_{k=0}^n \frac{m^k}{k!} = $$ $$= \sum_{k=0}^{3m} \frac{m^k}{k!} + \sum_{k=3m+1}^n\frac{m \cdot \ldots \cdot m}{(2m)(2m-1) \ldots1} \frac{1}{(2m+1) \ldots (3m) \ldots k} \leq $$ $$\leq \sum_{k=0}^{3m} \frac{m^k}{k!} + \sum_{k=3m+1}^n \left(\frac{1}{4} \right)^{k-2m} \leq \sum_{k=0}^{3m} \frac{m^k}{k!} + 1 < \infty$$
Now, an increasing and bounded sequence has a limit, proving the first claim.
I have been trying to see the product rule but could not finish it, at least using only the definition $e^t = \lim_n \left( 1+\dfrac tn\right)^n$. I have seen that: $$e^{s+t} = \lim_n \sum_{k=0}^n {n \choose k}\left(1+\dfrac sn\right)^{n-k}\left(\dfrac tn\right)^k \leq e^s \sum_{k=0}^n {n \choose k}\left(\dfrac tn\right)^k = e^se^t$$and this holds for $s \in \mathbb{R}$ and $t\geq 0$. On the other hand, if $s$ and $t$ are both negative, $$e^se^t = \lim_n \left( 1+\dfrac tn\right)^n\left( 1+\dfrac sn\right)^n= \lim_n \left( 1+\dfrac {t+s}n +\frac{st}{n^2}\right)^n \geq \lim_n \left( 1+\dfrac {t+s}n\right)^n=e^{s+t}$$Proving that, in general, $e^{s+t} \leq e^se^t$. Now, if $s$ and $t$ have opposite signs, the above inequaity changes its direction, and one has that $e^se^t \leq e^{s+t}$, so I still need to prove that $e^se^t \leq e^{s+t}$ when $s$ and $t$ are positive.
I hope this helps; this is not taken from any book, so probably there are faster ways to prove some facts.
EDIT: I finally got to prove it. Because $e^{s+t} = e^se^t$ holds for opposite signs, I get $e^{t}e^{-t}=1$, which is equivalent to $$\lim_n\left(1+\dfrac tn\right)^n = \lim_n \left(1-\dfrac tn\right)^{-n}$$And therefore $$\frac{e^se^t}{e^{s+t}} = \lim_n \left[ 1+\frac{st}{n(n-s-t)}\right]^{-n} \leq 1$$when $s$ and $t$ have the same sign.
Elementary results I am using:

*

*A bounded increasing sequence  has a limit

*Binomial theorem $(a+b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k}$

*If $x>-1$, $(1+x)^n \geq 1+nx$

*$\sum_{k=1}^n \frac{1}{4^k}\leq 1$ for any $n$
A: For brevity let's write $$f_n(t) =\left(1+\frac{t}{n}\right)^n\tag{1}$$ Using binomial theorem we can observe (after some algebra) that $$f_n(t) =1+t+\dfrac{1-\dfrac{1}{n}}{2!}t^2+\dfrac{\left(1-\dfrac{1}{n} \right)\left(1-\dfrac{2}{n}\right)}{3!}t^3+\dots\tag{2}$$ The series on right is finite and consists of $(n+1)$ terms. If $t>0$ then as $n$ increases each term in above series increases and the number of terms also increases. Hence $f_n(t) $ is an increasing sequence if $t>0$. The case $t<0$ will be handled later.
Let's first start with the simplest case when $t=1$ (note $t=0$ gives the limit $1$ as the sequence is constant with each term being $1$). We prove that $f_n(t) $ is bounded above by $3$ if $t=1$. Using equation $(2)$ we have $$f_n(1)\leq 1+1+\frac {1}{2!}+\dots+\frac{1}{n!}$$ which does not exceed $$1+1+\frac{1}{2}+\dots+\frac{1}{2^{n-1}}=1+2(1-(1/2^n))<3$$ Hence $f_n(1)$ converges to some limit and since $f_n(1)\geq 2$ the limit $f(1)$ lies in interval $[2,3]$. All we need to know is that $f(1)>0$.
Next let $t$ be a positive integer. We can write $$f_n(t) =\prod_{i=1}^{t}\dfrac{\left(1+\dfrac{1}{n+i-1}\right)^{n+i-1}}{\left(1+\dfrac{1}{n+i-1}\right)^{i-1}}\tag{3}$$ Each numerator above is a subsequence of $f_n(1)$ and hence tends to $f(1)$. Each denominator on the other hand tends to $1$ and thus $f_n(t) $ tends to $(f(1))^t$. Since the sequence is increasing it also follows that $$f_n(t) \leq (f(1))^t,\forall t, n\in\mathbb {N}\tag{4}$$ Now let's consider the case when $t>0$ is a real number. And let $m$ be a positive integer with $t<m$. Then using $(4)$ we have $$f_n(t) <f_n(m) \leq (f(1))^m$$ so that the sequence is bounded above and hence converges. Thus we have shown that $f(t) $ exists for all $t\geq 0$ and one can check $f(t) \geq 1+t$ and $f(t) =(f(1))^t$ if $t$ is a positive integer.
Consider $t$ as a positive rational number of the form $t=p/q$ where $p, q$ are positive integers. Then we know that $f_n(t) $ converges to $f(t) $. Further $$(f_n(t)) ^q=\left(1+\frac {p} {qn} \right) ^{qn} $$ is a subsequence of $f_n(p) $ and hence converges to $f(p) = (f(1))^p$. It follows that $(f(t)) ^q=(f(1))^p$. We have thus proved that $f(t)=(f(1))^t$ if $t$ is a positive rational number.
Let us now deal with $t<0$ so that $-t>0$. We can use Bernoulli's inequality and get $$1-\frac{t^2}{n}\leq\left(1-\frac {t^2}{n^2}\right)^n\leq 1\tag{5}$$ whenever $n>|t|$. By squeeze theorem it follows that $(1-(t^2/n^2))^n\to 1$ (alternatively it can be proved using a lemma of Thomas Andrews). In other words we have $$\left(1+\frac{t}{n}\right)^n=\dfrac {\left(1-\dfrac {t^2} {n^2}\right)^n} {\left(1+\dfrac{(-t)} {n} \right)^n} \to \frac{1}{f(-t)}$$ Thus we have proved that $f(t) $ exists for all real $t$ and we have $f(t) f(-t) =1$. This also proves that $f(t) >0$ for all real $t$.
To prove the exponential property of $f(t) $ we can consider the sequence $$a_n=\dfrac{1+\dfrac{s+t} {n} } {\left(1+\dfrac{s} {n} \right)\left(1+\dfrac{
t} {n} \right)} \tag{6}$$ It can be checked that $n(a_n-1)\to 0$ and hence by lemma of  Thomas Andrews the sequence $a_n^n\to 1$. Thus $f_n(s+t) $ converges to $f(s) f(t) $ and we have $f(s+t) =f(s) f(t) $ for all real $s, t$.
The case when $t\in\mathbb {C} $ is handled in this answer.
A: I have finally found a proof, but it is a long path. Too long to write it here.
First, I had to prove some simple facts like bounds of geometric sums, $\lim ka_n=k\lim a_n$, triangular inequality, etc.
Then I proved that the sequence is increasing and bounded for $t>0$. I needed to bound $\sum_{k=0}^n t^k/k!$ for this.
The hard part was proving that if $a_n$ is a sequence s. t. $a_n^n$ converges and $k\in\Bbb R$, then
$$\left(a_n+\frac k{n^2}\right)^n$$
converges to the same limit.
This allowed me state $e^te^{-t}=1$ (defining finally the function for $t<0$) and $e^te^s=e^{t+s}$.
The document has ten pages and it is written in Spanish.
A: My favorite proof i found in Best and Penner's Calculus.
The idea is to use a single interval  Riemann sum for the area under the curve  $1/x $ between  $1$ and $1+t/n $.
We get that $t/(n+t)\le\int_1^{1+t/n}1/x\rm dx\le t/n\implies e^{t/(n+t)}\le1+t/n\le e^{t/n}\implies e^{nt/(n+t)}\le (1+t/n)^n\le e^t$, and the result follows by letting  $n\to\infty $.
