I have a value where it needs to be decremented by $x\%$ successively $n$ times
For example, for
$$\begin{align*} \text{value} &= 100 \\ x &= 2\% \\ n &= 3 \end{align*}$$
we expect the value $94.1192$ because
$$\begin{align*} 100 - (\text{$2\%$ of 100}) &= 98\\ 98 - (\text{$2\%$ of 98}) &= 96.04\\ 96.04 - (\text{$2\%$ of 96.04}) &= 94.1192 \end{align*}$$
I am looking for a formula that can accomplish the goal.
On
Here's another direction:
Let $p$ be the percentage you decrement by $n$ times and let $a_0$ be the starting value.
In the example you give, you've essentially defined the first few terms of a sequence recursively. We can expand on that and define the sequence generally. We have $$ \begin{align} a_t&=a_{t-1}-\left(\frac{p}{100}\right)a_{t-1}\\ a_t&=a_{t-1}\left(1-\frac{p}{100}\right). \end{align} $$
You want to find a closed-form expression (a formula) for $a_n$ so that you don't have to calculate each term preceding the result you want.
The key is noticing that this is a geometric sequence with $(1-\frac{p}{100})$ as its common ratio. The closed-form of a geometric sequence is well-known: $$ a_n=a_0\left(1-\frac{p}{100}\right)^{n}. $$
Consider what it means to take $p\%$ off of an initial value $v_0$ to form a new value $v_1$:
$$v_1 = v_0 - (\text{$p\%$ of $v$}) = v_0 - \frac{p}{100} v_0 = \frac{100-p}{100} v_0$$
Take $p\%$ off of this again to form $v_2$:
$$v_2 = v_1 - (\text{$p\%$ of $v_1$}) = v_1 - \frac{p}{100} v_1 = \frac{100-p}{100} v_1$$
But then, we know $v_1$ in terms of just $v_0$:
$$v_2 = \frac{100-p}{100} \frac{100-p}{100} v_0 = \left( \frac{100-p}{100} \right)^2 v_0$$
Try this again and see if you can convince yourself of a pattern that gives $v_n$ for any positive integer $n$.