Wolfram says Sondow (2005)[1] and Sondow and Zudilin (2006)[2] noted the inequality:
$$\frac{1}{4rm}\left[\frac{(r+1)^{r+1}}{r^r}\right]^m < {(r+1)m \choose m} < \left[\frac{(r+1)^{r+1}}{r^r}\right]^m$$ for $m$ a positive integer and $r\ge 1$ a real number.

We can then use 
$${n+k-1\choose k} < {n+k\choose k} = {(r + 1)m \choose m}$$ with $r = \frac{n}{k}$ and $m = k$.

Then we have $${n+k-1\choose k} < \left[\frac{(r+1)^{r+1}}{r^r}\right]^m = \left(\frac{n+k}{k}\right)^{n+k}$$

Now, the binomial expression has the highest value at the middle of the Pascal's triangle. So, in our case, $n+k = 2k$ or at $k = n$.

Substituting that in the above inequality, we get:
$${n+k-1\choose k} < 2^{2n} = 4^n$$.

Therefore, a tighter bound is $${n+k-1\choose k} = O(4^n)$$.

You can also see that the lower bound for the maximum value is $${n+k-1\choose k} = \Omega\left(\frac{4^n}{n}\right)$$

References:  
[1] Sondow, J. "Problem 11132." Amer. Math. Monthly 112, 180, 2005.  
[2] Sondow, J. and Zudilin, W. "Euler’s constant, q-logarithms, and formulas of Ramanujan and Gosper" Ramanujan J. 12, 225-244, 2006.