Proof by induction that the sum of terms is integer I'm having some trouble in order to solve this induction proof.

Proof that $\forall{n} \in \mathbb{N}$ the number $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n$ is an integer. 

I've tried proving this by induction, but I've not succeed so far. What I did was:

Let's assume that $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$. So, by induction,  
$n=1$
  $\frac{1}{5} + \frac{1}{3} + \frac{7}{15} = 1$ and $1 \in \mathbb{Z}$.
Inductive step
  I want to prove that if $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$ then $\frac{1}{5}(n+1)^5+\frac{1}{3}(n+1)^3 + \frac{7}{15}(n+1) = l$, with $l \in \mathbb{Z}$

And then I'm stuck. I've tried developing every binomial but, for example, for $\frac{1}{5}(n+1)^5$ I get a $\frac{1}{5}n^4$ that is not in the inductive hypothesis, so therefore, I cannot get rid of it and it's not an integer, so I cannot conclude that $l \in \mathbb{Z}$.
Any help or ideas from where I can follow? Thanks in advance!

A direct solution (without using induction), is to realize that
$$\frac{n^5}{5} + \frac{n^3}{3} + \frac{7n}{15} = \frac{ 3n^5 + 5n^2 -8 } { 15} + 1 = \frac{ n (n-1)(n+1)(3n^2 + 8 ) } { 15} + 1 $$
$n(n-1)(n+1)$ is clearly a multiple of $3$ for all values of $n$.
$n(n-1)(n+1)$ is a multiple of $5$ if $n = 5k, 5k-1, 5k+1$.
$3n^2 + 8 $ is a multiple of 5 if $n = 5k+2, 5k+3$.
Hence, this is an integer.
