Help proving that $(n+a)^b = \Theta(n^b)$ Please you apologize me by my English.

I don't know how make that:
$$(n+a)^b = \Theta(n^b), b &gt; 0$$
I know, I must to find two constants such that:
$$ c_{1} n^b \leq (n+a)^b \leq c_{2} n^b $$

I do not know what else to do. I'v tried with the Newton's binomial, but I'm lost.
If $a \geq 0$ then 

$$(n+0)^b \leq (n+a)^b \,,$$

thus $c_1=1$ works.

Also, for all $n \geq a$ you have

$$(n+a)^b \leq (2n)^b =2^b n^b \,.$$

Now, fixing 

$$c_2 = \max \{ 2^b, \frac{(n+1)^b}{n^b},...,  \frac{(n+n-1)^b}{n^b} \}$$

you get the desired inequality.

For $a \leq 0$ you can get the inequlities the other way around, excepting that you'll have an issue if $a$ is a negative integer (what happens if $n=-a$?).
I proved it using the binomial expansion , which is

(x+y)n=nC0xny0+nC1xn−1y1+nC2xn−2y2+......nCnx0yn
now for the above problem (n+a)b=bC0nba0+nC1nb−1a1+nC2nb−2a2+......nCnn0an

we also now that for any polynomial
a0x1+a1x2+a2x2+⋯+anxn≤(a0+a1+a2+⋯+an)xn

substituting bC0a0
with C0, bC1a1 with C1

and so on (because that is a constant) we would get

(n+a)b=C0nb+C1nb−1+C2nb−2+......Cnn0

now i would jump to the following conclusion

C0nb≤C0nb+C1nb−1+C2nb−2+......Cnn0≤(C0+C1+C2+......Cn)nb

which makes (n+a)b=θ(nb)