The Binomial theorem for any index $n\in\mathbb{R}$ with $|x|<1,$ is
$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\ldots$
For $(x+a)^\pi$ one could take $x$ or $a$ common according as if $|a|<|x|$ or $|a|<|x|$ and use Binomial theorem for any index. i.e., $x^\pi(1+a/x)^\pi$ in case $|a|<|x|.$
Personally, I wouldn't have done it that way. So here is how I would've done it:
Method 1:
$$\sqrt2=\sqrt{1+1}=1+\frac12-\frac18+\dots\approx1+\frac12-\frac18=\frac{11}8=1.375$$
which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.
$$1.375^2=1.890625$$
Obviously it approaches the correct value as you take more terms.
Method 2:
This is called fixed-point iteration/Newton's method, and it basically goes like this:
$$x=\sqrt2\implies x^2=2$$
$$2x^2=2+x^2$$
Divide both sides by $2x$ and we get
$$x=\frac{2+x^2}{2x}$$
Now, interestingly, I'm going to call the $x$'s on the left $x_{n+1}$ and the $x$'s on the right $x_n$, so
$$x_{n+1}=\frac{2+(x_n)^2}{2x_n}$$
and with $x_0\approx\sqrt2$, we will then have $x=\lim_{n\to\infty}x_n$. For example, with $x_0=1$,
$x_0=1$
$x_1=\frac{2+1^2}{2(1)}=\frac32=1.5$
$x_2=\frac{2+(3/2)^2}{2(3/2)}=\frac{17}{12}=1.41666\dots$
$x_3=\dots=\frac{577}{408}=1.414215686$
And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$.
Best Answer
You are starting at the wrong end of the binomial. The first term should be $1^{15}$. This order is chosen because $x$ is much smaller than $1$, so the terms will decrease as you go along. To substitute $x=0.01$, just rewrite the term for that value. So if the exponent were $4,$, the first terms would be $1^4-4x^3+6x^2$ and substituting in $x=0.01$ would give $1-0.04+0.0006=0.9606$. The point is to see that the first few terms give a good approximation to the correct value, here $0.99^4=0.9659601$