Derivatives of exponential functions and number $e$ How to prove that this thing $ e = (1 + h)^\frac{1}{h}, h \rightarrow 0 \iff (1 + \frac n)^n, n \rightarrow \infty$
 goes to some exact value? Is there a proof of this, and if possible, intuition? (#)
If we want to find solution of equation $\frac{d}{dx} [a^x] = a^x$ we would easly see that solution is limit above, namely $e.$ But why is that? Is there intuitive reason why that golden value is, on a first sight, jut random irrational number? It's obvious that this 1 in limit is base-value, when time equals 0 ($e^0 = 1$). But I don't see conection in the rest of the formula (limit) :(
Also, I looked why $e^x = (1 + \frac{x}{n})^n, n \rightarrow \infty,$ (##) and here goes reasoning (I will always suppose that n goest to infinity):
Wee see that $e^{\frac{x}{n}} = 1.$ But also $1 + \frac{x}{n} = 1.$ Therefore, we get (##). Of course, this is just wrong: same "reasoning" can be done with any positive base. I must say that now I am confused: for very small $h$ we would have, for example when base is 3, $3^h = 1 + h$?? (###)
Can you prove (explain) questions above: (#), (###) and can you give me intuitive and clear picture of why we got that strange limit. I can get that number with algebra, but just can't with imagination and logic. 
 A: The magic of exponents is that doing addition in the input $$b^{x+y}$$ results in multiplication in the output $$b^x\times b^y$$.
As a result, an incremental change of $h$ will resultion in a change from $$x_0 \to x_0 + h$$ will result in an output change $$b^{x_0}\to b^{x_0}\times b^h$$ which is a proportion change based on the current value of $b^{x_0}$.
Now derivatives express a rate of change and with exponential powers that rate of change is proportional to the current value, so intuitively we should have, if $f(x) = b^x$ that $f'(x) = C_bb^x$ for some constant $C_b$.  And indeed we do:
$$f'(x) =\lim\limits_{h\to 0}\frac {f(x+h) - f(x)}h =\lim\limits_{h\to 0}\frac {b^{x+h} - b^x}h=\lim\limits_{h\to 0}\frac{b^x*b^h-b^x}h=b^x\lim\limits_{h\to 0}\frac{b^h-1}h = C_b b^x$$ where $C_b = \lim\limits_{h\to 0}\frac {b^h - 1}h$.
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Now if we think about this "increases is proportion to current" value is the the entire idea of interest.  You have $P(x_k)$ in principal at some time.  A period of time passe and you value increases by a factor of $r$ so you gain $r*P(x_k)$ and your new value is $P(x_{k+1}) = P(x_k) + r*(x_k)$.  And we recursively compund it to the formula that after $n$ units of time our initial investment of $P(0)$ will be $P(n) = P(0)\times (1+r)^n$.
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Now back to $f(x) = b^x$ and $f'(x) = b^xC_b$.  The larger the base, $b$, the quicker the rate of growth so the large $b$ is, the larger $C_b$ is.  The smaller $b$ is, the smaller $C_b$ is.  
Now if $a = 1$ then $f(x) = a^x = 1$ and $f'(x) = 1^x*C_1=0$ and $C_1 = 0$ and if $b$ is really huge there is no limit to how large $C_b$ can be.
So there must be some value $e$ where $C_e = 1$ and $f(x) = e^x$ so $f'(x) = e^xC_e = e^x*1 = e^x$.
So
$1=C_e = \lim\limits_{h\to 0}\frac {e^h-1}h$.  So we can estimate $e$ by solving the equation $\lim\limits_{h\to 0}\frac {e^h-1}h= 1$.
That is for teensy $h$.
$\frac {e^h-1}h \approx 1$
$e^h-1 \approx h$
$e^h \approx 1+h$
No lets replace teensy $h$ with $\frac 1n$ for big $n$.
$e^{\frac 1n} \approx (1+\frac 1n)$ and
$e \approx (1+\frac 1n)^n$.
And that's it:  $e =\lim\limits_{n\to \infty} (1+\frac 1n)^n$.
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Which.... if fits into the continuous compound interest aspect very nicely.
