Here's one way. Start with the expansion you want, using $a_0$, $a_1$, $a_2$, $a_3$, etc. for the unknown coefficients:

$$\sqrt{x+1}\;=\;a_{0}\;+\;a_{1}x\;+\;a_{2}x^2\;+\;a_{3}x^3\;+\;a_{4}x^4\;+\;a_{5}x^5\;+ ...$$

Finding $a_0$: Plugging in $x=0$ on both sides leads to $a_{0}=1$.

Finding $a_1$: Differentiate both sides of the expansion. This gives

$$\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}\;\;=\;\;a_{1}\;+\;2a_{2}x\;+\;3a_{3}x^2\;+\;4a_{4}x^3\;+\;5a_{5}x^4\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{1}=\frac{1}{2}$.

Finding $a_2$: Differentiate 2-times both sides of the expansion. This gives

$$\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{3}{2}}\;\;=\;\;2a_{2}\;+\;(2)(3)a_{3}x\;+\;(3)(4)a_{4}x^2\;+\;(4)(5)a_{5}x^3\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{2}=-\frac{1}{8}$.

Finding $a_3$: Differentiate 3-times both sides of the expansion. This gives

$$\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{5}{2}}\;\;=\;\;(2)(3)a_{3}\;+\;(2)(3)(4)a_{4}x\;+\;(3)(4)(5)a_{5}x^2\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{3}=\frac{1}{16}$.

Finding $a_4$: Differentiate 4-times both sides of the expansion. This gives

$$\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{7}{2}}\;\;=\;\;(2)(3)(4)a_{4}\;+\;(2)(3)(4)(5)a_{5}x\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{4}=-\frac{5}{128}$.

Finding $a_5$: Differentiate 5-times both sides of the expansion. This gives

$$\left(-\frac{7}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{9}{2}}\;\;=\;\;(2)(3)(4)(5)a_{5}\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{5}=\frac{7}{256}$.

Keep going to get as many coefficients as you want. If you keep careful track of the numbers without reducing the fractional expressions for the coefficients, you can easily determine a pattern (a pattern that can be proved by mathematical induction if you're so inclined).