You're close, but of course $np$ is not always an integer, which causes a problem.
Let $f(k) = {n \choose k} p^k (1-p)^{n-k}$ be the probability that there are $k$ successes in your experiment. Then look at the ratio $f(k)/f(k-1)$. You get:
$ {f(k) \over f(k-1)} = {{n \choose k} \over {n \choose k-1}} {p^k \over p^{k-1}} {(1-p)^{n-k} \over (1-p)^{n-k+1}} $
and cancelling powers of $p$ and $1-p$, this is
$ {{n \choose k} \over {n \choose k-1}} {p \over 1-p}. $
Now write the ratio of binomial coefficients in terms of factorials:
$ {{n \choose k} \over {n \choose k-1}} = {n! \over k! (n-k)!} {(k-1)! (n-k+1)! \over n!} = {n! \over n!} {(k-1)! \over k!} {(n-k+1)! \over (n-k)!} = {n-k+1 \over k}. $
Therefore, the original ratio we were interested in is
$ {f(k) \over f(k-1)} = {n-k+1 \over k} {p \over 1-p}. $
This decreases as $k$ increases. In particular, as $k$ increases it goes from being greater than $1$ to less than $1$, and so the sequence $f(0), f(1), \ldots, f(n)$ first increases and then decreases. (This is called being "unimodal".) Now, $f(k) \ge f(k-1)$ if and only if $ {n-k+1 \over k} {p \over 1-p} \ge 1 $ and we'll solve this inequality for k. Clear denominators: $ (n-k+1)p \ge k(1-p) $ Expand: $ np-kp+p \ge k-kp $ Cancel like terms: $ np+1 \ge k $ or $k \le np+p = (n+1)p$. So $f(k)$ increases until $k$ reaches $\lfloor (n+1)p \rfloor$ and then decreases.
For example, consider $k = 1/3, p = 6$. Our formula tells us that $f(k) \ge f(k-1)$ if and only if $k \le (6+1)/3$. So we'll have $f(2) \ge f(1)$ but $f(3) \le f(2)$; that is, 2 is th4e most likely number of successed. And indeed the probabilities $f(0), \ldots, f(6)$ are, in order, $0.088, 0.263, 0.329, 0.219, 0.082, 0.016, 0.001$, confirming the result.
But there's a special case when $(n+1)p$ is an integer. Then if $k = (n+1)p$ we actually get $f(k)/f(k-1) = 1$ -- so $(n+1)p$ and $(n+1)p-1$ are both maxima. The easiest example is when $p = 1/2$ and $n$ is odd -- say $n = 5$ -- then this predicts that $2$ and $3$ are both maxima, which they are. And of course if you flip five coins you expect two and three heads to be equally likely and more likely than any other number.
Since $f(0), f(1), f(2), \ldots, f(n)$ is increasing and then decreasing, the least likely number of successes is either $0$ or $n$. It's $0$ if $p > 1/2$ and $n$ if $p < 1/2$; if $p = 1/2$ then both are equally likely.