Recall that the derivative of the exponential function is `f^'(x) = e^x`. In fact, all the derivatives are `e^x`.

f '(0) = e0 = 1

f ''(0) = e0 = 1

f '''(0) = e0 = 1

We see that all the derivatives, when evaluated at x = 0, give us the value 1.

Also, `f(0) = 1`, so we can conclude the Maclaurin Series expansion will be simply:

`e^x~~1+x+1/2x^2` `+1/6x^3` `+1/24x^4` `+1/120x^5+...`

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