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Let

$f_n: \mathbb{R}^{+} \rightarrow \mathbb{R}$
If $x \leq n$ $f_n(x)=\left(1-\frac{x}{n}\right)^n$
Otherwise $f_n(x)=0$

I showed that its limit function is simly $e^{-x}$. Now I need to show that it converges uniformly on an interval $[0,\alpha]$ with $\alpha \in \mathbb{R}^{+}_{*}$. To do that, I figured I would study it's derivative, which is given by: $$f_n^{'}(x)= - \left(1 - \frac{x}{n}\right)^{n-1} $$

The derivative is of negative sign for $\forall x \leq n$, but $\forall x \geq n$, it is null( because the function itself is equal to zero if $x >n$.

As $n$ can be bigger or smaller than $\alpha$, I decided to divide the situation into two cases:

Case $\alpha $\forall x \in [0,n],\:\:\: f_n^{'}(x)<0$ and $\forall x, n\leq x<\alpha, f_n^{'}(x)=0$
I can conclude that $f_n(x)$ reaches it's maximum at $x=0$.

In the other case, I find the same conclusion.

Thus I can state that the max of $f_n$ is equal to $f_n(0)=1$.

Thus I have $||f_n -f||_{\infty}=0$

I have a somewhat doubtful feeling about my reasoning.

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    You've found the maximum of $f_n(x)$, but what you need (in fact, you don't) is to find maximum of $|f_n(x)-e^{-x}|$. – 2017-02-22
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    @SergeiGolovan What do you mean by "in fact, you don't"? Could I do something else? – 2017-02-22
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    There are a few applicable theorems for this case. 1) The Dini's theorem (you'd have to prove that the sequence is monotone first), 2) (Another theorem which connects monotonicity and uniform convergence)[http://math.stackexchange.com/questions/834126/sequence-of-monotone-functions-converging-to-a-continuous-limit-is-the-converge] – 2017-02-22

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