Like Qiaochu says, the key here is that the direct product of finitely many abelian groups functions as both the (categorical) product and the (categorical) coproduct in the category of abelian groups.
And before your eyes glaze over, what that means is that:
A homomorphism from an abelian group $A$ into a (finite) direct product $G_1\times G_2\times\cdots\times G_n$ of abelian groups is equivalent to a family $f_1,\ldots,f_n$ of homomorphisms $f_i\colon A\to G_i$; (in fact, this holds for arbitrary groups, and arbitrarily many direct factors, not just finitely many); and
A homomorphism from a finite direct product $G_1\times G_2\times\cdots\times G_n$ of abelian groups into an abelian group $B$ is equivalent to a family $g_1,\ldots,g_n$ of homomorphisms $g_i\colon G_i\to B$ (here we do need both finiteness and abelianness).
The first equivalence is easy: given a map $f\colon A\to G_1\times\cdots\times G_n$, the maps $f_i$ are just the compositions of $f$ with the canonical projections $\pi_i\colon G_1\times\cdots\times G_n\to G_i$; going the other way, given a family $f_1,\ldots,f_n$ of maps, you get the map $A\to G_1\times\cdots\times G_n$ by $f(a) = (f_1(a),f_2(a),\ldots,f_n(a))$.
For the second equivalence, given a homomorphism $g\colon G_1\times\cdots\times G_n\to B$, we define the maps $g_i\colon G_i\to B$ by restricting $g$ to the subgroup $\{0\}\times\cdots \times \{0\}\times B_i\times\{0\}\times\cdots\times\{0\}$. Conversely, given a family of homomorphisms $g_1,\ldots,g_n$, we construct the map $g$ by $g(x_1,\ldots,x_n) = g_1(x_1)+g_2(x_2)+\cdots+g_n(x_n)$; here, both the fact that the product has only finitely many factors and that the groups are abelian is important.
Now let $A=B=G_1\times\cdots\times G_n$. Then a homomorphism from $A$ to $B$ is equivalent, by 1, to a family of homomorphisms $\Phi_j\colon A\to G_j$. And by 2, each $\Phi_j$ is equivalent to a family of homomorphisms $\phi_{ij}\colon G_i\to G_j$. Thus, each homomorphism from $A$ to $B$ is equivalent to a family $\{\phi_{ij}\mid 1\leq i,j\leq n\}$, with $\phi_{ij}\colon G_i\to G_j$.
Now suppose you have two homomorphisms, $\Phi,\Psi\colon A\to B$, and you want to compose them. If $\Phi$ corresponds to $\{\phi_{ij}\}$ and $\Psi$ corresponds to $\{\psi_{ij}\}$, what does the composition correspond to in terms of maps $G_i\to G_j$?
If you trace the correspondence carefully, you should find that the induced map from $G_i$ to $G_j$ is precisely
$$\psi_{i1}\circ\phi_{1j} + \psi_{i2}\circ\phi_{2j}+\cdots+\psi_{in}\circ\phi_{nj},$$
so that if you arrange the families $\{\phi_{ij}\}$ and $\{\psi_{ij}\}$ into matrices, composition corresponds to matrix multiplication in the usual way (though because composition is not commutative, you have to be mindful of the order of the products.
Once you have that endomorphisms can be "coded" as matrices with composition corresponding to matrix multiplication, the fact that automorphisms correspond to invertible matrices follows immediately. However, actually writing down a formula is complicated, because these matrices have entries that don't commute with one another; even in simple cases, like trying to do something like $C_{p^{\alpha}}\times C_{p^{\beta}}$ with $\alpha\gt\beta$, writing down the inverse of an automorphism in terms of its entries turns into a computation with congruences that is difficult to write down as a formula. But never fret, you aren't asked for an explicit formula.
Let $|G_{i}|=p_{i}^{n_{i}}$ for all $i$. Let $(g_{1},\ldots,g_{r})\in\ker\phi$. Then $g_{1}\cdots g_{r}=1$. If $r=1$, then we are done. If $r\geq 2$, for any $i$, we have $g_{i}=(g_{1}\cdots g_{i-1}g_{i+1}\cdots g_{r})^{-1}$. Since $g_{i}^{p_{i}^{n_{i}}}=1$, $|g_{i}|\mid p_{i}^{n_{i}}$. Since
$$
g_{i}^{p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}}}=(g_{1}\cdots g_{i-1}g_{i+1}\cdots g_{r})^{-p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}}}=1,
$$
$|g_{i}|\mid p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}}$. So $|g_{i}|\mid \gcd(p_{i}^{n_{i}},p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}})=1$. Then $g_{i}=1$. So $\phi$ is injective.
Best Answer
Recall the following criterion:
Criterion. A group $G$ is a direct product if and only if it has two normal subgroups $H,K$ such that $H \cap K = \{1\}$ and $G = HK$.
This criterion gives, by induction
Criterion bis. A group $G$ is a $n$-fold product if and only if it has $n$ normal subgroups $H_1,\ldots,H_n$ such that $G = H_1 \ldots H_n$ and $H_i \cap (H_1 \ldots H_{i-1} H_{i+1} \ldots H_n) = \{1\}$.
In you case you have the Sylow subgroups $P_1,\ldots,P_r$ which are normal by hypothesis. The equality $G = P_1 \ldots P_r$ follows from cardinality considerations, and $P_i \cap (P_1 \ldots \widehat{P_i} \ldots P_n) = \{1\}$ follows from order considerations.
Indeed, you can change the order in the product! In fact, let $g \in G_i$, $h \in G_j$. Then $g h g^{-1} h^{-1} \in G_j \cap G_i$. But $G_j$ is a $p_j$-group and $G_i$ is a $p_i$-group! If an element lies in their intersection, its order must divide both $p_i^n$ and $p_j^m$, hence it is $1$.