What is the formula for diminishing percentages?

$0.1$ 1 times is $0.1$
i.e.: $x$
$0.1$ 2 times is $−0.1^2+2*0.1 = 0.19$
i.e.: $(1-x)*x+x$
$0.1$ 3 times is $0.1^3-3*0.1^2+3*0.1 = 0.271$
i.e.: $(1-((1-x)*x+x))*x+(1-x)*x+x$
$0.1$ 4 times is $-0.1^4+4*0.1^3-6*0.1^2+4*0.1 = 0.3439$
i.e.: $(1-((1-((1-x)*x+x))*x+(1-x)*x+x))*x+(1-((1-x)*x+x))*x+(1-x)*x+x$

How can I find the function of this where $f(n)$ gives 0.1 $n$ times?
IE: $f(1) = 0.1$ | $f(2) = 0.19$ | $f(3) = 0.271$ | $f(4) = 0.3439$

Further illustrated in words:

1x: You have a 10% chance to do something.
2x: You have a 19% chance to do something.
3x: You have a 27.1% chance to do something.
4x: You have a 34.39% chance to do something.

Credits to @Arthur for the correct information which led to finding the answer. Instead of looking for an Exponential Decay (increasing form) it was much easier to invert an Exponential Decay function. The answer to my inquiry was:

$$Y=1-(0.9^x)$$

This successfully returns what I was searching for. IE:

X = 2
$Y=1-(0.9^2)$
$Y=1-(0.81)$
$Y=0.19$

X = 4
$Y=1-(0.9^4)$
$Y=1-(0.6561)$
$Y=0.3439$